THE    ELEMENTS 


OF 


PHYSICAL  CHEMISTRY 


BY 


J.   LIVINGSTON    R.    MORGAN,    PH.D., 

M 

Professor  of  Physical  Chemistrv,  Columbia  University. 


THIRD    EDITION,    REVISED   AND   ENLARGED. 
FIRST    THOUSAND. 


UN1VERG 

OF  S 


NEW  YORK: 

JOHN  WILEY    &    SONS. 

LONDON  :   CHAPMAN   &   HALL,   LIMITED. 

1905. 


Copyright,  1899,  1902,  1905, 

BY 
J.  L.  R.  MORGAN. 


ROBERT  DRUMMOND,    PRINTER,    NEW   YORK. 


PREFACE  TO  THE   FIRST   EDITION. 


THE  object  of  this  book  is  to  present  the  elements 
of  the  entire  subject  of  Physical  Chemistry  in  one  volume, 
together  with  the  important  and  but  little  known  appli- 
cations of  it  to  the  other  branches  of  chemistry.  Many 
persons  have  found  it  difficult  to  obtain  a  comprehensive 
outline  of  the  subject,  owing  to  the  length  of  time  which 
it  has  been  necessary  to  spend  upon  the  separate  volumes 
devoted  to  special  portions  of  it.  To  all  such  this  volume 
may  be  of  value. 

It  is  especially  intended  as  a  text-book  for  either  class- 
work  or  self-instruction,  and  although  the  calculus  is 
used  in  the  derivation  of  some  of  the  laws,  still  much 
can  be  done  without  any  training  in  the  higher  mathe- 
matics. In  general,  references  are  given,  so  that  any 
one  wishing  to  make  an  extended  study  of  any  special 
portion  may  do  so  with  little  difficulty.  The  amount 
of  the  subject  included,  however,  embraces  that  which 
is  likely  to  be  useful  to  all  chemists,  and  is  that  which 


vi  PREFACE   TO  THE  SECOND  EDITION. 

to  those  studying  alone,  as  well  as  to  others  who  have 
already  studied  the  subject  but  not  yet  attempted  to 
apply  it. 

J.  L.  R.  M. 

HAVEMEYER  LABORATORIES, 
December,  1901. 


PREFACE  TO  THE  THIRD   EDITION. 


As  a  basis  for  the  revision  of  this  work  I  have  been  so 
fortunate  as  to  have  a  copy  of  the  former  edition  contain- 
ing suggestions  and  criticisms  from  the  hand  of  Prof. 
Ostwald.  For  his  great  kindness  in  having  volunteered 
to  take  this  trouble,  and  for  many  other  favors,  both 
during  my  student  days  and  since,  I  am  indeed  very 
grateful,  and  I  cannot  allow  this  opportunity  to  pass 
without  expressing  to  him,  as  far  as  words  may  suffice, 
my  sincerest  thanks.  Further,  I  wish  to  acknowledge 
the  influence  of  his  "  Vorlesungen  uber  Naturphilosophie," 
some  points  of  which  I  have  attempted  to  apply  in  this 
edition. 

The  ideas  I  have  had  in  mind  in  making  this  revision 
may  be  summarized  as  follows:  To  bring  the  subject- 
matter  up  to  date;  to  distinguish  sharply  between 
hypothesis  and  fact,  avoiding  the  former  as  far  as  is 
possible;  and  to  accentuate  the  physical  meaning  of 

the  results  of  mathematical  reasoning,  i.e.,   to  employ 

vii 


Vlii  ,    PREFACE   TO   THE    THIRD  EDITION. 

mathematics  simply  as  a  means  to  an  end,  and  to  make 
the  matter  as  intelligible  as  possible,  even  to  the  non- 
mathematical  reader.  In  few  words  the  motive,  if  I 
may  so  express  myself,  of  this  edition  is  to  treat  the 
subject  as  something  which  can  be  applied  to  other 
branches  and  to  illustrate  the  application  and  methods 
of  application  by  the  use  of  numerous  problems.  In 
this  way  it  is  hoped  that  the  subject,  instead  of  being 
merely  a  more  or  less  interesting  theoretical  study,  as 
is  too  often  the  case,  may  appeal  to  the  student  as  a  tool 
by  the  aid  of  which  actual  results  may  be  obtained.  This 
plan  is  the  expression  of  the  need  of  my  own  students, 
and  I  assume  consequently  that  it  will  be  welcome  to 
others  who  not  only  wish  to  know  the  subject,  but  also 
to  use  it. 

J.  L.  R.  M. 

COLUMBIA  UNIVERSITY, 
May,  1905. 


CONTENTS. 


CHAPTER   I. 

PACE 

INTRODUCTORY  REMARKS i 

i.  Physical  chemistry.  2.  Energy.  3.  The  factors  of 
energy.  4.  Atomic  and  molecular  weights. 

CHAPTER   II. 

THE  GASEOUS  STATE 9 

5.  Definition  of  a  gas.  6.  The  gas  laws.  7.  The  specific 
gravity  of  gases.  8.  Methods  of  determining  the  specific  grav- 
ity. 9.  Abnormal  vapor-densities.  Dissociation.  10.  Vol- 
ume, pressure,  and  concentration,  n.  Variation  from  the  gas 
laws.  The  equation  of  Van  der  Waals.  12.  Specific  heat. 
The  first  principle  of  thermodynamics.  13.  Determination  of 
the  specific  heat  of  gases.  14.  The  second  principle  of  thermo- 
dynamics. 15.  The  cycle.  Entropy.  1 6.  "The  factors  of  heat 
energy. 

CHAPTER   III. 

THE  LIQUID  STATE 61 

17.  Distinction  between  liquids  and  gases.  18.  Connection 
between  the  gaseous  and  liquid  states.  19.  Vapor-pressure  and 
boiling-point.  20.  The  heat  of  evaporation.  21.  The  relation 
between  vapor-pressure,  heat  of  evaporation,  and  temperature. 
22.  Refraction  of  light.  23.  Surface-tension.  24.  Molecular 
vreight  in  the  liquid  state.  Critical  temperature. 


CONTENTS. 


CHAPTER   IV. 

PAGE 

THE  SOLID  STATE 91 

25.   Remarks.     26.   Atomic  heat.     Law  of  Dulong  and  Petit. 
27.  Changes  in  the  state  of  aggregation. 


CHAPTER   V. 

THE  PHASE  RULE 104 

28.  Object  of  the  phase  rule.  29.  The  phase  rule.  30.  Deri- 
vation of  the  phase  rule.  31.  The  equilibrium  of  water  in  its 
phases.  32.  The  phase  rule  and  the  identification  of  basic  salts. 


CHAPTER    VI. 

SOLUTIONS 116 

33.  Definition  of  a  solution.  34.  Gases  in  liquids.  35.  Liq- 
uids in  liquids.  36.  Solids  in  liquids.  37.  Osmotic  pressure. 
38.  Electrolytic  dissociation  or  ionization.  39.  Solution-pres- 
sure. 40.  Vapor-pressures  of  solutions.  41.  The  relation  be- 
tween osmotic  pressure  and  the  depression  of  the  vapor-pressure. 
42.  Increase  of  the  boiling-point.  43.  Depression  of  the  freez- 
ing-point. 44.  Division  of  a  substance  between  two  non-misci- 
ble  solvents.  Depressed  solubility.  45.  Solid  solutions.  46.  Col- 
loidal solutions.  47.  The  molecular  weight  in  solution. 


CHAPTER  VII. 

THERMOCHEMISTRY 200 

48.  Definition.  49.  Application  of  the  principle  of  the  con- 
servation of  energy.  50.  The  heat  of  formation.  51.  Chemical 
changes  at  a  constant  volume.  52.  Chemical  changes  at  a  con- 
stant pressure.  53.  Relation  between  results  for  constant  volume 
and  constant  pressure.  54.  The  effect  of  temperature.  55.  The 
thermal  reactions  of  electrolytes. 


CONTENTS.  xi 

CHAPTER   VIII. 

PAGE 

CHEMICAL  CHANGE. 

A.  Equilibrium 222 

56.  Reversible  reactions.  57.  The  law  of  mass  action. 
58-  Equilibrium  in  homogenous  gaseous  systems.  59.  Equi- 
librium in  non-homogenous  systems.  60.  Dissociation  of  a 
solid  into  more  than  one  gas.  61.  Equilibrium  in  liquid  systems. 
62.  The  effect  of  temperature  upon  an  equilibrium.  The  varia- 
tion of  the  constant  of  equilibrium  with  the  temperature. 

B.  Chemical  Kinetics 261 

63.  Application  of  the  law  of  mass  action.  64.  Reactions  of 
the  first  order.  65.  Catalytic  action  of  hydrogen  ions.  Catal- 
ysis. 66.  Reactions  of  the  second  order.  67.  Reactions  of 
the  third  order.  68.  Incomplete  reactions.  69  Reactions  be- 
tween solids  and  liquids.  70.  Speed  of  reaction  and  temperature. 

C.  Application  of  the  Law  of  Mass  Action  to  Electrolytes — Ionic 
Equilibria 281 

71.  Organic  acids  and  bases.  The  Ostwald  dilution  law. 
72.  Acids,  bases,  and  salts  which  are  ionized  to  a  considerable 
extent  Empirical  dilution  laws.  73.  Heat  of  ionization. 
^jta.  74.  Solubility  or  ionic  product.  75.  Hydrolytic  dissociation. 
Hydrolysis.  76.  Determination  of  the  ionization  constant  from 
observations  of  increased  solubility.  77.  Ionic  equilibria. 
78.  The  color  of  solutions.  79.  The  action  of  indicators. 
80.  General  analytical  reactions. 


CHAPTER    IX. 

ELECTROCHEMISTRY. 

A.  The  Migration  of  the  Ions 364 

81.  Electrical  units.  82.  Faraday's  law.  83.  The  migra- 
tion of  the  ions.  84.  Determination  of  the  relative  velocity  of 
migration. 

B.  The  Conductivity  of  Electrolytes 374 

85.  The  specific  conductivity.     86.  Molecular  and  equivalent 

conductivity.  87.  Determination  of  electrical  conductivity. 
88.  Ionic  conductivities.  General  rules.  89.  The  conductivity 
of  organic  acids.  90.  The  absolute  mobility  of  the  ions. 


xii  CONTENTS. 

PAGE 

91.  The  basicity  of  an  acid.  92.  The  conductivity  of  neutral 
salts.  93.  The  ionization  of  water.  94.  The  temperature 
coefficient  of  conductivity.  95.  Conductivity  of  difficultly  solu- 
ble salts.  96.  The  dielectric  constant  and  dissociating  power. 
Other  solvents  than  water.  97.  The  conductivity  of  mixtures 
of  substances  having  an  ion  in  common. 

C.  Electromotive  Force 399 

98.  Determination  of  the  electromotive  force.  99.  Types  of 
cells.  100.  Chemical  or  thermodynamical  theory  of  the  cell. 
101.  The  osmotic  theory  of  the  cell.  102.  Mathematical  ex- 
pression of  the  osmotic  theory  of  the  cell.  103.  The  measure- 
ment of  the  potential  difference  between  a  metal  and  a  solution. 
104.  The  heat  of  ionization.  105.  Concentration  cells.  106. 
Dissociation  by  aid  of  the  electromotive  force.  107.  Electrolytic 
solution  pressures.  108.  Cells  with  inert  electrodes.  109. 
Processes  taking  place  in  the  cells  in  common  use. 

D.  Electrolysis  and  Polarization .  440 

no.  Decomposition  values,  in.  -Theory  of  polarization. 
112.  Primary  decomposition  of  water.  113.  Electrolytic  sepa- 
ration of  metals  by  graded  electromotive  forces. 


CHAPTER   X. 

PROBLEMS    .  453 

TABLES o 485 

APPENDIX 497 

INDEX .  503 


ELEMENTS    OF    PHYSICAL: ^  CHEMISTRY. 


CHAPTER  I. 
INTRODUCTORY  REMARKS. 

i.  Physical  Chemistry  is  that  branch  of  the  science 
oj  chemistry  which  has  for  its  object  the  study  of  the  laws 
governing  chemical  phenomena. 

Other  titles  for  this  subject  are  also  in  use  (General 
Chemistry,  Theoretical  Chemistry),  but,  since  the  sub- 
jects treated  lie  in  that  border-land  between  Physics 
and  Chemistry,  and  since  many  purely  physical  methods 
are  used,  the  term  Physical  Chemistry  is  the  most  de- 
scriptive one,  and  is  to  be  preferred.  The  subject  of 
Physical  Chemistry  is  usually  divided  into  two  parts: 
Stoichiometry  and  Chemical  Energy,  the  latter  term  in- 
cluding the  laws  of  affinity  and  all  other  allied  subjects. 
As  it  is  difficult,  however,  to  make  a  sharp  distinction, 
we  shall  consider  both  together,  dividing  the  subject 
only  into  minor  portions  in  the  form  of  chapters. 


2  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

2.  Energy. — Since  much  of  our  work  has  to  do  with 
the  different  forms  of  energy,  it  will  be  well  first  to  recall 
some  points  which  later  we  shall  use  constantly. 

Force  is  whatever  changes,  or  tends  to  change,  the 
motion  of  a  body  by  altering  its  direction  or  its  magni- 
tude* :  The  -  unit .  of  force  is  the  dyne.  A  dyne  acting 
upon  a  gram,  for  one  second  will  give  it  the  velocity 
of  i  centimeter  per  second,  and  n  dynes,  n  centimeters 
per  second.  At  Washington  a  body  falling  freely  for 
one  second  acquires  the  velocity  of  980.10  centimeters, 
therefore  the  force  of  gravitation  of  i  gram  at  Wash- 
ington is  equal  to  980.1  dynes. 

Work. — The  unit  of  work  is  that  work  which  is  done 
when  unit  force  is  overcome  through  unit  distance.  This 
is  called  the  erg.  We  have  then 

dynes  X  centimeters  =  ergs. 

The  maximum  work  to  be  obtained  from  any  process  is 
that  amount  which  can  be  produced  under  ideal  con- 
ditions. 

Energy  is  work  or  anything  which  can  be  transformed 
into  work,  or  produced  from  work.  Mechanical  energy  is 
of  two  kinds,  kinetic  and  distance — the  former  being  due 
to  motion,  the  latter  to  position.  As  all  forms  of  energy 
can  be  transformed  the  one  into  the  other,  the  general 
unit  of  energy  is  the  erg.  Since  it  is  impossible  to  re- 
move all  the  energy  from  a  body,  we  have  no  way  of 


INTRODUCTORY  REMARKS.  3 

determining  how  much  is  contained  in  it.  We  can  only 
measure  the  excess  of  energy  which  a  body  contains  in 
a  given  state  over  that  which  it  contains  in  a  certain 
standard  state.  This  is  determined  by  allowing  a  body 
to  go  from  the  given  state  to  the  standard  state  in  such 
a  way  that  the  difference  in  energy  all  appears  in  a 
measurable  form,  for  example,  as  heat.  If  we  measure 
this  heat,  we  have  the  energy  of  the  body  in  the  given 
state,  as  compared  to  that  in  the  normal  state. 

3.  The  factors  of  energy. — Experience  has  shown  that 
the  possibility  of  a  transfer  of  energy  does  not  depend 
upon  the  amount  of  the  energy  in  question,  but  rather 
upon  the  intensity  of  the  energy.  For  this  reason  the 
amount  of  energy  is  usually  expressed  as  a  product  of 
two  factors.  The  one  factor  determines  the  amount  of 
energy  which  can  and  must  be  absorbed  by  a  body  in 
going  from  one  state  into  another.  This  is  called  the 
capacity  jactor  of  the  energy.  Upon  the  other  factor 
depends  the  possibility  of  the  transfer  of  the  energy 
from  one  body  to  another  in  such  a  way  that  if  the  value 
of  this  factor  for  both  the  bodies  is  the  same  no  transfer 
takes  place.  This  is  called  the  intensity  j actor  of  the 
energy.  Bodies  witK  the  same'  value  for  the  intensity 
factor  are  in  equilibriufe,  i.e.,  no  finite  transfer  of  energy- 
takes  place  between  them.  If  the  intensity  factor  has  a 
different  value  in  two  states  of  a  substance,  an  exchange 
of  energy  will  take  place  between  them  until  the  intensity 


4  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

factor  has  become  of  the  same  value  for  both  and  equi- 
librium is  established.  In  general  the  capacity  factor 
can  be  distinguished  from  the  intensity  factor  by  the 
fact  that  in  all  changes  the  sum  of  the  capacity  factors 
is  a  constant,  which  is  not  true  for  the  intensity  factor. 

Examples. — For  heat  energy  the  capacity  factor  is 
called  entropy,  while  the  temperature  is  the  intensity 
factor.  If  two  bodies  at  the  same  temperature  are 
brought  in  contact,  there  is  no  change  in  them.  Should 
the  temperatures  be  different,  however,  there  is  an  exchange 
of  heat  energy  of  such  a  nature  that  the  temperatures 
become  the  same.  Bodies  containing  different  amounts 
of  heat,  when  brought  together,  remain  unchanged  pro- 
vided their  temperatures  are  the  same.  The  old  unit  of 
heat  energy,  the  calorie  (18°),  is  equal  to  41,830,000  ergs. 

Volume  energy  is  equal  to  the  work  required  to  cause  a 
decrease  of  volume,  or  that  which  is  done  by  an  increase 
of  volume.  The  intensity  factor  is  the  pressure,  while 
the  volume  is  the  capacity  factor.  Imagine  a  gas  enclosed 
in  a  cylinder  which  is  provided  with  a  movable  partition. 
If,  on  each  side  of  this  partition,  we  have  the  gas  under 
different  pressures,  then  the  partition  will  move  to  the 
side  with  the  smaller  pressure,  until  the  two  become 
equalized.  As  long  as  the  pressure  is  the  same  the 
volumes  may  have  any  relation  without  causing  the  parti- 
tion to  move.  Pressure  is  expressed  in  dynes  per  square 
centimeter,  and  its  unit  is  that  pressure  under  which  an 


INTRODUCTORY  REMARKS.  5 

increase  of  volume  equal  to  i  cubic  centimeter  will  do 
the  work  equal  to  i  erg.  The  pressure  of  the  atmos- 
phere, then,  is  equal  to  76X980.1  Xi3. 5953=  1,012,681, 
i.e.,  a. little  over  a  million  of  these  units. 

For  kinetic  energy,  equal  to  i/2Mv2,  the  capacity 
factor  is  the  mass,  and  the  intensity  factor  is  the  square 
of  the  velocity.  i/2Mv2y  it  is  true,  may  also  be  divided 
into  the  factors  Mv(= momentum)  and  v(  =  velocity). 
In  this  case  the  momentum  is  the  capacity  factor,  the 
velocity  the  intensity  factor. 

For  electrical  energy  the  amount  of  electricity  is  the 
capacity  factor,  the  electromotive  force  being  the  intensity 
factor. 

In  the  same  way  all  other  energies  may  be  divided 
into  factors,  and  it  has  been  found  that  the  study  of  the 
different  energies  is  much  simplified  by  the  process. 

In  general,  then,  for  all  energies  we  have 

(i)  E=ci, 

where  c  is  the  capacity  factor  and  i  the  intensity  factor, 
E  representing  the  energy  itself. 

If  the  energy  is  increased  by  an  infinitesimal  amount, 
then 

dE  =  d(ci)  =  cdi  +  idc; 

and  from  this  we  may  find  the  equations  for  c,  i,  dcy 
and  di.  They  are: 


6  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

7  t1 

(2)  c  =  -j-r(dc=o,  c  =  const.); 

(3)  di = — (dc = o,  c  =  const.) ; 

(4)  •i=-r-(di  =  o,  i  =  const.); 

dE 

(5)  dc=*—r-(di—o,  i  =  const.). 


If  in  a  system  there  are  two  kinds  of  energy  which  are 
so  related  that  a  change  in  the  one  causes  a  corresponding 
change  in  the  other,  i.e.,  one  is  a  .function  of  the  other, 
then,  if  the  two  kinds  of  energy  are  E\  and  £2, 

dE1=dE2 
or 

d(ciii)=d(c2i^\  • 
i.e.,  if  c= const. 
(6)  Cidii  =  c2di2. 

This  method  of  deriving  an  equation  giving  the  relation 
between  two  kinds  of  energy  in  one  system  is  very  useful, 
and  we  shall  have  occasion  to  use  it  later. 

4.  Atomic  and  molecular  weights. — Since  we  are  to 
make  very  extensive  use  of  these  two  factors  in  our  later 
work,  it  is  quite  necessary  here  to  make  clear  the  exact 


INTRODUCTORY  REMARKS.  7 

meanings  of  the  words,  as  we  shall  employ  them.  A 
glance  at  the  practical  application  of  the  words  shows 
that  there  is  nothing  hypothetical  about  them,  other  than 
their  derivation.  It  is  unfortunately  true,  however,  that 
from  this  one  early  acquires  the  idea  that  the  actual 
results  of  an  analysis  or  chemical  reaction  are  dependent 
not  only  upon  the  practical  work,  but  also  upon  the  molecu- 
lar or  atomic  hypothesis.  This  idea  is  so  strong,  indeed, 
that  it  is  not  uncommon  to  find  people  assuming  that  any 
direct  proof  contrary  to  the  hypothesis  would  actually 
affect  the  practical  results  themselves.  It  is  not  possible 
to  discuss  this  question  in  detail  here,  so  the  reader  must 
be  referred  elsewhere.  The  definition  of  the  two  terms, 
however,  will  serve  to  show  that  they  are  simply  expres- 
sions of  experimentally  determined  facts,  and  in  such  a 
sense  they  are  used  throughout  this  book. 

The  combining  weight  of  any  element  is  the  weight 
which  combines  with  16  units  of  weight  of  oxygen,  or  some 
multiple  or  submultiple  of  that;  or  combines  with  the 
combining  weight  of  some  other  element,  the  value  of 
which  is  expressed  hi  terms  of  oxygen.  The  combining 
weight  is  identical  with  the  atomic  weight,  when  there 
is  but  one  combining  ratio;  and  is  usually  the  least 
common  multiple  of  the  combining  weight  in  case  there 
is  more  than  one  combining  ratio. 

The  molecular  or  formula  weight  in  the  gaseous  state 
(that  in  other  states  will  be  defined  later)  expresses  the 


8  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

number  of  grams  of  gas  which  at  o°  C.  and  i  atmosphere 
pressure  occupy  the  same  volume  as  two  combining 
weights  of  oxygen  (16X2  =  32  gr.),  i.e.,  approximately 
22.4  liters  (see  p.  12),  or  a  corresponding  volume  at 
any  other  temperature  or  pressure.  Since  the  factor  for 
the  conversion  of  grams  into  ounces  (av.)  (0.0353)  *s 
the  same  as  that  for  the  conversion  of  liters  into  cubic 
ieet,  this  relation  also  holds  for  ounces  (av.)  and  cubic 
feet.  In  other  words,  the  molecular  weight  of  any  gas, 
expressed  in  ounces  (av.),  occupies  the  volume  of  22.4 
cubic  feet. 


CHAPTER  II. 
THE   GASEOUS   STATE. 

5.  Definition  of  a  gas. — A  gas  is  distinguished  by  its 
property  of  indefinite  expansion.     In  other  words,  a  gas 
is  limited  in  volume  only  by  the  walls  of  the  vessel  which 
contains  it. 

6.  The  gas  laws. — These  are  the  result  of  experience 
as  to  the  behavior  of  gases  in  general  under  varying 
conditions. 

Boyle's  (Mariotte's)  law:  At  constant  temperature  the 
volume  oj  any  gas  is  inversely  proportional  to  the  pres- 
sure under  which  it  exists. 

If  v  represents  the  volume  of  any  weight  of  a  gas  at 
the  pressure  p,  and  v\  the  volume  of  the  same  amount 
of  the  gas  at  the  pressure  pi,  then 

(7)        pv  =  piVi=  constant  for  constant  temperature. 

If  the  temperature  is  different  for  the  two  cases  the 
product  pv  is  no  longer  equal  to  the  product  p\v-\_. 

The  effect  of  a  change  in  temperature  upon  the 
volume  of  a  gas  is  given  by  the  law  oj  Charles  (Dalton, 
Gay-Lussac);  The  volume  of  any  gas  increases  by  the 

9 


io  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

1/273  Part  °l  MS  volume  at  o°  C.  jor  every  increase  o]  Us 
temperature  equal  to  i°  C. 

If  the  temperature  were  decreased  continuously  from 
o°,  we  would  finally  reach  a  point  at  which  the  volume 
is  equal  to  zero.  Provided  that  the  law  of  Charles  holds 
for  such  a  low  temperature,  this  point  will  be  273  centi- 
grade degrees  below  o°  C.  This  point  is  called  the 
absolute  zero,  and  the  temperature  measured  from  it  in 
centigrade  degrees  is  the  absolute  temperature.  This  is 
designated  by  the  letter  T,  the  temperature  according 
to  the  centigrade  scale  being  distinguished  by  the  letter 
t.  Between  these  two  terms,  then,  we  have  the  relation 

r=/  +  273,     or    *=r-273. 

The  volume  of  a  gas  is  thus  proportional  to  its  abso- 
lute temperature,  provided  its  pressure  remains  constant; 
and,  by  Boyles'  law,  its  pressure  is  proportional  to  the 
absolute  temperature,  when  its  volume  remains  con- 
stant. From  this  we  can  now  find  the  value  of  the 
product  pv  for  any  temperature.  We  have 

vocT(p  =  const.); 


-(r  =  const.); 


hence 

T 

v«f 


THE  GASEOUS  STATE.  1 1 

and 

jT 

v  =  k— 
P 

or 

(8)  pv  =  kT. 

But  at  o°  C. 

(80)  pov0  =  k2  73, 

and  combining  (8)  and  (So),  with  elimination  of  the  con- 
stant k,  we  find 


273 


This  term  —  —   is  a  constant  for  any  one  gas,  how- 


ever, for  />o  =  iO33  grams  per  square  centimeter  (i.e., 
76X13.6)  and  ^o  is  the  volume  occupied  by  a  definite 
weight  of  the  gas  under  this  pressure  at  the  tempera- 
ture of  o°  C. 

If  vQ  is  the  volume  of  i  gram  of  gas  at  o°,    76  cms., 
we  have 


where  v  is  the  volume  of  i  gram  of  gas  at  the  tempera- 


12  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

ture  T  and  the  pressure  p,  'and  r  is  the  specific  gas  con- 

i 
slant  oj  the  gas. 

Or,  comparing  such  quantities   of  different  gases   as 
will  give  the  volume  of  22.4  liters  at  o°  C.  and  i  atmos- 
phere pressure  (i.e.,  comparing  the  molecular  weights), 
we  find 
(9)  pV=RT(\.z.,  ^°—  for  i  mole*  =  12  J, 

where  R  is  the  molecular  gas  constant,  which  is  the  same 
for  all  gases,  r  multiplied  by  the  molecular  weight  of 
the  gas,  then,  must  be  equal  to  R. 

Equation  (9)  is  the  equation  oj  state  for  gases.  In  it 
p,  the  pressure,  is  to  be  expressed  as  a  weight  in  grams 
upon  the  square  centimeter,  V  is  the  volume  of  one 
mole  in  cubic  centimeters,  and  T  is  the  absolute  tem- 
perature in  centigrade  degrees,  i.e,,  T  =  1  +  273°. 

The  molecular  gas  constant,  R,  may  be  calculated  as 
follows  : 


In  tf-,    Fo  =  22400   cc.,  hence   ^2*400X1033 

273  273 

=  84800,  where  the  weight  of  i  gram  per  square  centi- 
meter is  taken  as  the  unit.  To  transform  this  into  abso- 
lute units  it  is  then  only  necessary  to  multiply  it  by 
980.1  (page  2). 

*  From  here  on  We  shall  call  the  molecular  Weight  in  grams  the  mole, 
as  has  been  proposed  by  Prof.  Qstwald.  A  mole  of  any  gas  at  o°  and 
76  cms.  pressure  occupies  22.4  liters,  which  is  the  average  result  of  many 
determinations. 


THE   GASEOUS  STATE.  13 

Or,  if  po  is  given  in  atmospheres  and  FO  in  liters,  we 
have 

R=-7p—= — -  =0.0821  liter-atmospheres, 

and       R=    ^    =8.3iXio7  ergs,  when   p   is    given   in 

dynes  and  V  in  cc. 

Dalton's  law  refers  to  the  pressure  exerted  by  the 
single  gases  in  a  mixture  of  gases. 

The  pressure  exerted  upon  the  walls  of  a  vessel  con- 
taining a  mixture  oj  gases  is  equal  to  the  sum  of  the  pres- 
sures which  the  single  gases  would  exert  were  they  alone 
in  the  vessel.  The  exact  meaning  of  this  will  be  seen 
from  the  following  example:  When  into  a  closed  ex- 
hausted vessel  we  introduce  i  gram  of  gas  it  will  exert 
a  certain  pressure  upon  each  square  centimeter  of  the 
surface.  If  now  we  introduce  a  second  gram  of  the 
same  or  a  different  gas,  it  will  exert  exactly  the  same 
pressure  upon  the  vessel  that  it  would  have  exerted  had 
the  first  gram  not  been  there.  Upon  the  walls,  however, 
with  the  same  gas,  the  pressure  will  be  doubled. 

One  law  still  remains  to  be  considered  which  has  had 
and  still  has  great  value  in  chemistry;  it  is  Avogadro's 
law  or  hypothesis:  All  gases  under  the  same  conditions 
oj  pressure  and  temperature  contain  in  unit  volume  the 
same  number  oj  gram  molecules.  In  its  original  form 
this  relation  referred  to  the  hypothetical  molecules  them- 


14  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

selves  (not  to  gram  molecules),  and  hence  was  hypothet- 
ical. In  the  form  given-  above,  however,  it  is  simply 
expressive  of  the  definition  of  molecular  weight  already 
given  (pp.  7,  8).  As  a  matter  of  fact  the  original  form 
has  always  been  used  of  late  as  though  it  did  refer  to 
gram  molecules,  so  that  the  hypothetical  character  has 
been  lost,  and  the  relation  may  well  be  designated  as 
Avogadro's  law.  The  advantage  of  the  statement  in 
this  form  is  that  it  accentuates  the  difference  between 
hypothesis  and  fact,  and  shows  plainly  that  the  relation 
itself  is  free  from  all  hypothesis. 

By  determining  the  density  of  any  gas  referred  to 
oxygen  it  is  very  easy,  then,  to  determine  its  molecular 
or  formula  weight.  Since  equal  volumes  under  like 
conditions  contain  the  same  number  of  moles,  the  weights 
of  these  volumes  will  be  related  as  the  molecular  weights, 
the  basis  of  which  is  that  of  oxygen,  i.e.,  32.  The  term 
formula  weight,  which  we  shall  use  as  identical  with 
molecular  weight,  is  to  be  preferred  to  the  latter,  for  the 
formula  shows  distinctly  the  relation  between  atomic 
weight  (i.Q.,  the  combining  ratio)  and  molecular  weight 
(i>e.,  the  number  of  combining  or  atomic  weight  which 
occupy  the  volume  of  22.4  liters  at  o°  C.  and  760  mm. 
Hg  pressure).  Thus  H  represents  the  atomic  weight 
of  hydrogen  (i.e.,  1.008),  while  H2  is  the  molecular  or 
formula  weight  (i.e.,  2X1.008  gr.);  and  for  compounds 
this  is  much  more  convenient,  especially  since  the  mole- 


THE  GASEOUS  STATE.  15 

cular  or  formula  weight  is  not  a  fixed  quantity  for  all 
conditions,  but  varies  with  these.  -  Thus  we  have  FeCls 
and  Fe2Cl6,  according  to  the  conditions,  and  a  glance 
at  the  formula  shows  at  once  the  molecular  weight. 

7.  The  specific  gravity  of  gases.  —  The  specific  gravity 
of  a  gas  is  the  weight  of  the  unit  of  volume,  but  as  this 
is  always  very  small  it  is  customary  to  find  in  these  terms 
the  value  of  a  certain  gas  taken  as  a  standard,  and  then 
to  express  the  density  of  other  gases  in  terms  of  this  gas, 
taken  as  unity. 

As  the  composition  of  air  varies,  oxygen  has  been 
chosen  as  the  chemical  standard.  Since  this  is  very 
easily  obtained  in  the  pure  state  and  its  combining 
weight  with  other  elements  can  be  accurately  determined, 
its  advantages  as  a  standard  are  apparent. 

One  liter  of  O  weighs  1.4290  grams  at  o°  and  76 
centimeters  pressure.  Its  specific  volume,  i.e.,  the 
volume  of  i  gram  at  any  temperature  and  pressure,  is 


_ 


0.0014290 

Its  density,  i.e.,  the  weight  of  i  cc.  at  any  temperature 
and  pressure,  is  the  reciprocal  of  the  specific  volume, 
i.e., 

d  =  -  =0.0014290  —  —  -7  grams. 

v  -   76(1+1/275)  G 


i6  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

The  molecular  volume,  since  02  =  32,  is 


If  w  is  the  weight  of  a  gas,  and  g*is  the  weight  of  an 
equal  volume,  under  like  conditions,  of  the  standard 
gas,  i.e.,  O,  the  specific  gravity  of  the  gas  is 


w 


The  relation  in  weight  between  hydrogen  and  oxygen 
is  as  1:15.88;  between  hydrogen  and  air  is  as  1:14.48. 
In  this  way  knowing  the  density  referred  to  one,  that 
referring  to  either  of  the  others  can  be  determined. 
Referred  to  air  the  density  can  be  employed  to  find  the 
molecular  weight  by  aid  of  the  formula 

M  =  0.001  293X2  2400^  =  28.96^, 

where  0.001293  is  the  weight  of  i  liter  of  air  at  o°  and 
76  cm.  Hg  pressure,  and  A  is  the  density  of  the  gas  based 
upon  air  as  a  standard. 

8.  Methods  of  determining  the  specific  gravity.  — 
Here  we  shall  consider  the  subject  both  for  gases  and 
for  the  vapors  generated  from  substances  by  heat,  since 
the  determination  of  the  latter  is  of  importance  for  the 
ascertaining  of  the  molecular  weight.  For  convenience 
we  shall  divide  the  subject  into  three  groups  of  methods, 


THE  GASEOUS  STATE.  17 

but  shall  consider  only  the  theory  of  the  methods,  assuming 
that  the  student  will  look  up  the  practical  details  in  one 
of  the  many  laboratory  manuals. 

A.  The  weight  oj  a  certain  'volume  of  the  gas. — This 
may  be  used  both  for  gases  and  for  the  vapors  given  off 
by  substances  under  high  temperatures. 

a.  For  gases. — The  weight  of  a  certain  volume  is 
found  by  weighing  first  a  balloon  filled  with  the  gas, 
and  then  exhausting  this  and  weighing  it  alone.  In 
this  way  we  can  find  the  weight  of  the  volume  of  gas, 
and  by  comparing  this  with  the  weight  of  the  same  volume 
of  oxygen  ascertain  the  density. 

P.  For  gases  generated  from  substances  by  heat  (Dumas) . 
— Here  the  process  is  slightly  different.  The  substance 
is  placed  in  a  weighed  flask  with  a  slender  neck  and  the 
whole  heated  at  a  constant  temperature  until  the  sub- 
stance is  all  transformed  into  gas,  when  the  neck  of  the 
flask  is  fused  together.  After  cooling,  the  flask  is  weighed 
and  the  neck  cut  off  so  that  the  volume  of  the  flask  can 
be  found  from  the  weight  of  water  it  will  contain.  We 
have  then  the  weight  of  a  certain  number  of  cubic  centi- 
meters of  the  gas,  which,  when  compared  with  the  weight 
of  an  equal  vohime  of  oxygen  at  that  temperature,  will 
give  the  density  referred  to  that  gas,  and  then  by  calcu- 
lation that  based  upon  hydrogen  or  air. 

This  method  of  Dumas  has  been  further  modified, 
so  that  it  is  possible  to  work  under  diminished  pressure 


*8  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

and  consequently  at  lower  temperatures,  which  is  of 
great  importance  in  case  the  substance  decomposes  at  a 
higher  temperature.  The  flask  for  this  purpose  is  the 
same  as  that  used  before  except  that  the  neck  is  con- 
nected with  an  air-pump  and  manometer.  When  the 
pressure  has  reached  the  desired  point  the  temperature 
is  increased  until  the  substance  goes  into  the  gaseous 
form,  when,  as  above,  the  neck  is  sealed.  The  calcula- 
tion is  the  same  as  before  except  that  the  pressure  under 
which  the  gas  exists  is  equal  to  that  of  the  barometer 
minus  that  of  the  manometer. 

B.  The  volume  occupied  by  a  certain  weight. — By 
these  methods  we  avoid  the  error  which  is  always  present 
when  the  weight  of  a  volume  of  gas  is  determined.  The 
substance,  in  solid  or  liquid  form,  is  weighed  and  then 
transformed  into  the  gaseous  state,  the  volume  which 
it  then  occupies  being  measured.  The  method  of  Gay- 
Lussac,  as  improved  by  Hofmann,  is  intended  only  for 
the  determination  of  the  specific  gravity  of  the  gases 
formed  from  solid  or  liquid  substances.  A  weighed 
amount  of  substance  is  brought  into  the  Torricelli  vacuum 
of  a  barometer-tube,  which  is  surrounded  by  a  tempera- 
ture bath,  and  the  volume  of  the  gas  measured.  The 
gas  is  under  a  pressure  equal  to  that  of  the  atmosphere 
minus  that  of  the  column  of  mercury  in  the  tube.  This 
volume  of  gas,  reduced  to  o°  C.  and  76  cms.  of  Hg,  weighs 
just  what  the  original  substance  did,  and  this  weight 


THE  CASEOUS  STATE.  19 

compared  with  that  of  an  equal  volume  of  the  standard 
gas  will  give  the  density. 

The  method  of  Victor  Meyer  differs  from  that  of 
Hofmann  in  that  the  gas  is  formed  in  a  flask  and  displaces 
the  air,  the  volume  of  this  at  a  lower  temperature  being 
measured  in  a  burette  connected  with  the  flask.  This 
volume  is  of  course  the  same  as  that  the  substance  itself 
would  assume  at  that  temperature,  provided  that  no  con- 
densation takes  place,  for  the  coefficient  of  expansion  of 
all  gases  is  the  same. 

The  method  under  this  group  which  may  be  used  for 
gases  in  general  is  to  pass  a  determined  volume  of  the 
gas  over  or  through  a  weighed  substance  which  will 
absorb  it  all.  In  this  way  the  difference  in  weight  of  the 
substance  before  and  after  the  passage  of  gas  gives  the 
weight  of  the  determined  volume  of  gas.  Thus  oxygen 
may  be  absorbed  by  glowing  metallic  copper,  the  copper 
oxide  formed  giving  the  weight  of  copper  plus  oxygen. 

C.  The  value  of  this  method  (Bunsen's),  which  is 
adapted  only  to  gases,  is  that  but  very  small  amounts 
of  substance  are  needed  for  the  experiment.  It  depends 
upon  the  relation  of  the  velocity  of  outflow  of  a  gas 
through  a  small  aperture  to  the  specific  gravity  of  the  gas. 
The  formula  is  derived  as  follows:  Let  p  be  the  differ- 
ence between  the  pressure  under  which  a  gas  exists  and 
that  of  the  atmosphere,  and  v  be  the  volume  of  the  gas 
which  flows  out  in  the  unit  of  time.  The  total  work  done 


20  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

by  the  change  in  volume  is  then  pv.  Since  this  work  is 
used  to  force  the  gas  out,  it  is  equal  to  the  increase  oi 
kinetic  energy  of  the  gas,  i.e., 


where  c  is  the  velocity  of  outflow  of  the  gas,  and  m  is 
its  mass.  If  equal  volumes  of  different  gases  are  consid- 
ered, flowing  through  the  same-sized  openings  under  the 
same  pressure,  then 


or 


or,  since  the  masses  of  equal  volumes  are  proportioned 
to  their  densities 


The  apparatus  used  by  Bunsen  is  very  simple.  A 
glass  tube  having  a  very  fine  opening  at  one  end,  below 
which  is  a  stop-cock,  is  clamped  into  a  vessel  of  mercury. 
In  this  tube  is  a  piece  of  glass  rod  to  act  as  a  float,  while 
on  the  outside  there  are  two  marks  close  together.  The 
gas  is  now  passed  into  the  tube  at  a  certain  pressure, 
and  the  tube  lowered  into  the  mercury  until  the  float  is 
just  at  the  lower  mark.  The  cock  is  then  opened  and  the 
time  necessary  for  the  passage  of  the  float  from  the  lower 


THE  GASEOUS  STATE.  21 

to  the  upper  mark  observed.     Since  the  time  is  inversely 
proportioned  to  the  velocity,  we  have 


By  carrying  out  the  experiment  for  one  gas  and  then 
for  oxygen  we  can  find  the  density  of  the  gas  in  terms 
of  oxygen.  The  results  by  this  method,  and  this  is 
true  in  a  lesser  degree  for  all  methods  for  molecular 
weight,  are  not  exact,  but  are  simply  intended  to  indicate 
what  number  of  combining  weights  of  an  element,  or 
what  weight  of  a  compound,  represents  the  formula 
weight.  For  this  reason  very  great  accuracy  is  not 
necessary,  for  it  is  only  a  question  of  finding  whether 
the  combining  weight  or  the  simplest  ratio,  as  found 
by  analysis,  is  to  be  multiplied  by  the  factor  i,  2,  3,  etc. 

9.  Abnormal  vapor  densities.  Dissociation.  —  The 
specific  gravity  or  density  of  the  gases  generated  by  heat 
from  substances  is  found  in  many  cases  to  be  too  small, 
i.e.,  the  molecular  weights  thus  determined  are  smaller 
than  the  theoretical  values.  And,  further,  the  molecular 
weight  is  found  to  vary  with  the  temperature,  pressure 
and  some  other  factors.  These  results,  naturally,  must 
be  obtained  with  greater  accuracy  than  those  mentioned 
above  if  the  process  is  to  be  followed  at  all  closely. 
Since,  according  to  our  definition  of  molecular  weight, 
i  mole  occupies  22.4  liters  at  o°  and  760  mm.  Hg 
pressure,  a  smaller  density  would  represent  a  decom- 


22  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

position  of  the  substance  employed.  Thus  if  a  substance 
decomposes  in  such  a  way  that  from  each  molecule  of 
the  gas  there  are  two  others  formed,  then  the  vapor 
density  must  be  one  half  what  it  should  be.  For  each 
volume  of  the  undecomposed  gas  we  shall  have  two 
volumes  of  the  gases  formed  from  it  at  the  same  tem- 
perature, since  by  Avogadro's  law  we  have  the  same 
number  of  molecules  in  equal  volumes.  These  two 
volumes,  however,  will  weigh  the  same  as  the  original 
one  volume,  so  that  the  weight  of  one  volume  of  the 
mixed  gases  will  be  one  half  that  of  one  volume  of  the 
undecomposed  gas.  St,  Clair  Deville  called  this  de- 
composition dissociation.  Thus  NKUCl  dissociates  ac- 
cording to  the  scheme 


where  the  sign  <=±  means  that  the  reaction  may  go  in 
either  direction,  according  to  the  conditions.  That 
these  two  gases,  NHs  and  HC1,  are  actually  present  in 
the  vapor  of  NH4C1  can  be  shown  in  the  following  way 
(Pebal  and  Than)  :  A  lump  of  solid  NH4C1  is  placed  in 
a  tube  upon  an  asbestos  plug,  and  the  temperature  of 
the  tube  raised.  The  NH4C1  now  volatilizes  and  dis- 
sociates in  NHa  and  HC1.  Since  NHa  is  the  lighter 
gas,  it  diffuses  more  rapidly  through  the  asbestos  plug 
than  the  HC1,  consequently  on  one  side  "of  the  plug  we 
shall  have  an  excess  of  HC1  and  on  the  other  an  excess 


THE   GASEOUS  STATE.  23 

of  NH3.  The  presence  of  these  two  products  may  be 
shown  by  passing  a  current  of  dry  air  through  the  parts 
of  the  tube  on  each  side  of  the  plug  and  then  over 
moistened  litmus  paper,  which  will  show  the  nature  of 
the  gases  present. 

Since  the  density  of  the  substance,  in  case  of  a  dis- 
sociation, decreases  to  an  extent  dependent  upon  the 
temperature  it  is  an  important  thing  to  be  able  to  find 
the  degree  of  the  dissociation  at  any  one  temperature, 
i.e.,  to  determine  the  extent  of  the  reaction  (p.  22)  form- 
ing the  products  on  the  right. 

This  can  be  found  from  the  relation  of  the  vapor 
density  to  the  dissociation,  i.e.,  the  dependence  of  both 
upon  the  number  of  moles.  If,  for  example,  a  is  the 
percentage  of  the  gas  dissociated,  i.e.,  the  degree  of  disso- 
ciation, and  we  start  with  i  mole  of  the  gas,  then 
i— a  is  the  undissociated  portion.  If  there  are  n  moles 
of  the  products  formed  from  i  mole  of  the  gas,  the  total 
number  of  moles  present  at  any  time  is 

(i  -  a)  +  na  =  i  +  (n  —  i)a. 

The  ratio,  then,  of  i  to  i  +  (n—  i)a  will  be  the  same 
as  that  of  the  vapor  density  as  it  is,  to  the  vapor  density 
as  it  should  be,  i.e., 


24  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Where  du  is  the  vapor  density  as  it  should  be  (i.e., 
as  it  is  without  dissociation),  and  dd  is  what  it  actually 
is.  The  degree  of  dissociation,  then,  is 

du  —  dd 


(n-i)dd 

If  a  substance  dissociates  completely  into  two  products, 
its  vapor  density  is  1/2  what  it  should  be;  if  into  three, 
1/3,  etc. 

Examples. — NH4C1  v.d= nearly  1/2  what  it  should  be; 
hence 

NH4C1  = 


NH2CO2NH4*;.</  =  nearly  1/3;  hence 
NH2C02NH4  =  C02  +  2NH3. 

When  i  mole  of  gas  is  formed  from  a  solid  or  liquid 
against  the  pressure  p,  the  work  done  is  equal  to 
pV(=RT).  In  the  case  the  substance  gives  off  a  gas 
which  dissociates  into  two  or  more  gases,  i  mole  of 
the  original  substance  is  equal  to  more  than  i  mole 
of  gas,  and  the  external  work  done  is  equal  to  pV=iRTt 
where  i  is  the  total  number  of  moles  of  gas,  equal  to 
(i  —a)  -\-noi.  The  value  of  pV,  since  it  may  be  the  same 

for  any  number  of  single  values  of  p  and  V  (for  #<*—), 

can   be   most   easily   and   simply   determined   from   the 
other  side  of  the  equation,  i.e.,  from  iRT,  which  has 


THE   GASEOUS  STATE.  2$ 

but  one  value  for  any  temperature.  The  energy  terms 
obtained  in  such  a  case  will  then  depend  solely  upon 
the  choice  of  the  units  in  which  R  is  expressed,  for  T  is 
a  simple  ratio. 

The  process  of  dissociation  is  a  very  common  one, 
aryd  proofs  of  it,  quite  as  good  as  the  one  by  Pebal  and 
Than  already  mentioned,  are  numerous.  A  few  of  them 
are  given  briefly  below. 

PCls  when  heated  shows  a  green  color  which  increases 
with  the  temperature.  This  is  due  to  the  dissociation 
into  PC13  and  C12. 

N2C>4  by  heat  becomes  brownish,  red,  due  to  the  for- 
mation of  2NG>2.  At  500°  C.  this  color  disappears  en- 
tirely, for  then  2NC>2  breaks  down  into  2NO  and  O2, 
both  of  which  are  colorless. 

When  a  gas  composed  of  a  heavy  metallic  and  a  light 
gaseous  element  is  dissociated  by  heat  in  an  open  tube, 
and  cooled  suddenly,  the  heavy  element  will  crystallize 
out.  This  is  due  to  the  fact  that  the  two  gases  go 
through  the  tube  with  a  different  velpcity  and  the  sud- 
den cooling  forms  a  compound  of  what  is  left,  and  the 
heavy  element,  being  in  excess,  remains  to  a  certain 
extent  in  the  free  state.  An  example  of  this  is  given 
by  Marsh's  test. 

The  specific  heat  of  a  dissociating  gas  is  very  large 
owing  to  the  fact  that  the  heat  supplied  increases  the 
dissociation  as  well  as  the  temperature.  When  the 


26  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

dissociation  is  complete,  however,  the  specific  heat  be- 
comes constant.     Thus  we  have  for  acetic  acid  gas: 

/ 129°         160°         200°         240°         280° 

Sp.  h't 1.50        1.27        0.95        0.64        0.480 

The  specific  heat  of  a  dissociating  gas  also  varies 
greatly  with  the  pressure,  as  does  the  dissociation. 

The  conduction  of  heat  by  a  dissociating  gas  is  very 
much  larger  than  by  a  gas  not  capable  of  dissociating. 
The  heat  at  the  one  side  causes  dissociation,  and  the 
products  of  dissociation  go  to  the  colder  portion,  where 
they  unite,  giving  up  heat  and  causing  other  molecules 
to  dissociate,  which  in  their  turn  diffuse  and  unite,  etc. 
This  process  continues  until  the  gas  is  all  dissociated, 
when  its  conduction  of  heat  becomes  normal. 

Experiment  shows  that  a  dissociation  takes  place  to 
a  smaller  extent  when  one  of  the  products  of  the  dissocia- 
tion is  already  present.  Thus  PCls  is  almost  entirely 
undissociated  in  presence  of  an  excess  of  chlorine  or 
PCls,  i.e.,  its  vapor  density  under  such  conditions  is 
found  to  be  104.5  m  place  of  the  calculated  value  104. 

Later,  under  Chemical  Change,  we  shall  find  a  formula 
giving  the  relations  between  the  original  and  final  sub- 
stances in  a  chemical  reaction  for  a  constant  temperature, 
and  another  showing  the  effect  of  a  change  in  temperature, 
but  here  it  is  simply  necessary  to  note  that  dissociation 
depends  upon  temperature,  pressure  and  presence  of 


THE  GASEOUS  STATE.  27 

the  products  of  the  dissociation.  The  following  tables 
will  serve  to  show  how  great  the  effect  of  temperature 
and  pressure  is  upon  the  dissociation  of  a  gaseous 
substance: 

DISSOCIATION  OF  NITROGEN  TETROXIDE,  N2O4. 
(Density  of  N2O4  =  3.18;  of  NO2+NO2=i.59;  air  =i.) 


Temp.                  Sp.  Gr.  of  Gas. 

Percentage  Dis 

26°.  7                      2.65 

19.96 

35°-  4                       2.53 

25-65 

39°.  8                       2.46 

29.23 

49°.  6                       2.27 

40.04 

60°.  2                       2.08 

52.84 

70°.  o 

.92 

65-57 

80°.  6 

.80 

76.61 

90°.  o 

.72 

84-83 

100°.  I 

.68 

89-23 

in0.  3 

•65 

92.67 

121°.  5 

.62 

96.23 

135°-° 

.60 

98.69 

154°.° 

•58 

IOO.OO 

DISSOCIATION  OF  PC15. 
(Density  PC16  =  7.2;  PC13+C12=3.6;  air-i.) 


Temp. 
182° 
190° 
200° 
2300 
250° 
274° 
288° 
300° 


Density.          Percentage  Dissociation. 

5.08  41.7 

4-99  44-3 

4-85  48.5 

4.30  67.4 

4.00  8o.O 

3-84  87.5 

3.67  96.2 

3-65  97-3 


28  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

DISSOCIATION  OF  N2O4. 

(Equal  Temperatures,  Varying  Pressures.) 

Temp.  Pressure.  Density  (air  =  i). 

i8°.o  279       mm.  2.71  17.3 

i8°.s  136         "  2.45  29.8 

20°. O  301  "  2.70  17.8 

20°. 8  153.5     "  2.46  29.3 

The  molecular  weights  (see  definition,  p.  7)  of  some  of 
the  elements  in  the  gaseous  state  are  given  below  and 
will  serve  to  show  the  importance  of  the  process  of  dis- 
sociation, and  how  dependent  the  molecular  weight  is 
upon  the  temperature. 


ARSENIC. 

t°C. 

M. 

644 

309 

As4  =  3oo 

670 

308 

As2=-isc 

1715 

X57 

1736 

160 

PHOSPHORUS. 

313 

128 

500 

126.1 

P4=I24 

1484 

105 

P2=    62 

1678 

93-3 

1708 

9i 

IODINE. 

253 

285 

1330 

162 

We  see  from  these  that  the  higher  the  temperature 
the  lower  the  molecular  weight.  Iodine  at  high  tempera- 
tures is  monatomic,  as  are  mercury  between  446°-!  731°, 
cadmium  at  1040°,  tin  at  1400°,  and  sodium  and  potassium. 


THE  GASEOUS  STATE.  29 

And  sulphur  in  the  gaseous  state,  according  to  the 
temperature,  may  be  S$,  S*i  or  $2,  the  latter  being  the 
condition  at  800°  and  above.  Mixtures  of  these  three 
forms  can  also  exist,  at  the  lower  temperatures,  the 
molecular  weight  of  these  lying  between  the  two  extremes, 
Ss  and  £2;  at  higher  temperatures,  1900-2000°  C.,  the 
•$2  is  observed  to  break  down  further  into  the  form  of 
25  (Nernst),  although  Biltz  and  Meyer  found  the 
formula  weight  to  be  63.5  at  1719°  C. 

Among  compounds  we  find  the  chlorides  of  iron  and 
aluminium,  Fe2Cl6  and  A^Cle  only  to  exist  at  com- 
paratively low  temperatures,  being  present  in  the  forms 
FeCla  and  Aids  at  higher  ones. 

10.  Volume,  pressure  and  concentration.— Thus  far 
we  have  only  used  density  to  show  the  number  of  gram 
molecules  present  at  any  time.  It  is  also  possible,  how- 
ever, to  express  this  in  other  ways.  The  V  in  equation 
(9),  it  will  be  remembered,  was  used  to  designate  the 
number  of  liters  of  space  in  which  i  mole  exists,  and 
consequently  the  volume  of  a  gas  or  of  a  mixture  of 
gaseous  substances  will  give  the  number  of  moles  present, 
provided  the  temperature  and  pressure  be  known. 

Since  volume  and  pressure  for  any  one  temperature 
are  inversely  proportional,  the  number  of  moles  can 
also  be  determined  by  the  pressure  at  a  known  tempera- 
ture and  volume.  At  times  density,  volume' and  pressure 
can  be  employed  with  equal  facility;  at  others,  however, 


30  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

one  of  these  is  often  found  to  be  simpler  in  application 
than  the  others.  This  term  concentration  is  used  to  express 
the  number  oj  moles  oj  substance  per  liter,  and  is  designated 
throughout  this  work  by  the  letter  c.  ^ 

The  relations  of  these  quantities  to  one  another  is  not 
difficult  to  express.  Since  the  concentration  of  i  mole 
per  liter  at  o°  gives  the  pressure  22.4  atmospheres,  and 
a  pressure  at  other  temperatures  which  is  proportional 
to  the  absolute  temperature,  concentrations  (pressures) 
can  be  transformed  readily  in  pressures  (concentrations). 
And  the  same  is  true  of  volumes  under  constant  pressure. 
As  it  is  quite  necessary  to  become  familiar  with  the  use 
of  these  terms  (and  especially  in  their  application  to 
dissociation  phenomena)  in  our  later  work,  the  following 
examples  are  given: 

At  190°  C.  the  reaction  PC15  <r>  PC13  +  C12  goes  to  the 
right  to  such  an  extent  that  44.3%  of  the  original  quantity 
of  PICs  is  decomposed,  i.e.,  a,  the  degree  of  dissociation, 
is  0.443.  Starting  with  i  mole  of  PC15  at  190°  and 
atmospheric  pressure,  and  assuming  that  no  dissociation 
takes  place,  the  volume  occupied  by  the  gas  would  be 


22.4  ---  liters.     Since  we  lose  0.443  °f  tne  penta- 

chloride  and  gain  a  like  fraction  of  a  mole  of  the  tri- 
chloride and  of  chlorine,  the  final  number  of  moles  will 
be  equal  to  (1-0.443)4-2X0.443,  or  1.443.  As  the 
volume  (at  constant  temperature  and  pressure)5  is  pro- 


THE  GASEOUS  STATE.  31 

portional  to  the  number  of  moles  present,  and  this  has 
gone  from  i  to  1.443,  the  final  volume  will  be  1.443 

times   the   original   volume,    i.e.,    1.443X22.4    • 

liters.  In  case  the  volume  remains  constant,  i.e.,  the 
internal  pressure  increases  (assuming  that  the  increase 
of  pressure  has  no  effect  upon  the  dissociation)  the 
pressure-change  will  be  equal  to  the  volume-change 
when  the  pressure  remains  constant.  Since  the  original 
(i.e.,  undissociated)  volume  of  the  mole  of  PC15  at  190° 

273  +  100   , 
is    22.4     —  -  -     liters   at   atmospheric   pressure,    the 

final  pressure  in  the  volume  of  i  liter  will  be  I.443X 
22.4  —  —  -  atmospheres.  Of  this  value,  1.443,  the 

/  O 


fraction  1—0.443  *s  due  to  tne  Pdo>  °-443  to  PCla  and 
0.443  to  tne  chlorine.     The  partial  pressures  at  equili- 

270  -\-  IQO 
brium,  then,  are    0.557X22.4  ---  atmospheres  of 

PCls,  and  0.443X22.4  --  each  for  the  trichloride 
•  -73 

and  the  chlorine  (Dalton's  law). 

To  express  the  quantity  of  these  three  substances  in 
terms  of  concentration  (i.e.,  in  moles  per  liter)  it  is  only 
necessary  to  divide  the  final  number  of  moles  of  each 
substance  by  the  total  volume  of  the  system.  Thus 
of  the  mole  of  PCls  with  which  we  started  we  have  but 
1—0.443  mole  at  equilibrium,  and  gain  0.443  mole  of 


32  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

each  of  the  two  constituents.      Since  the  final  volume  is 

2*7  3  +  IOO 

1.443X22.4  — —   -  liters,  the  final  concentrations  are 


273 
1-0.443 


273  +  I9° 
1.443X22.4-^^ 

Q-443 

273  +  190 
1.443X22.4-^- 


:CPC13  =  ^chlorine. 


Here,  naturally,  the  volume  changes.  In  case  the 
volume  remains  constant  it  is  to  be  treated  in  the 
same  way,  for  it  is  still  the  final  volume.  In  the  above 

273+190 
case  it  would  be  22.4 . 

Example. — 10  gr.  of  PC^  at  atmospheric  pressure  (in 
a  cylinder,  for  instance,  which  is  provided  with  a  movable 
piston)  are  heated  to  250°  C.,  at  which  temperature  a 
is  equal  to  0.8.  Find  the  partial  pressure  and  concen- 
tration of  each  of  the  three  gases  present  at  equilibrium. 

If  the   PC\5  were  undissociated  at  this  temperature 

10             273  +  2^0    , 
the     10    gr.  would    occupy  -~ —  22.4 liters, 

V  l£         *o  /  o 

v» 

280.3  being  the  molecular,  molar,  or  formula  weight  of 
the  PCls.     Since  a  =  0.8,  at  equilibrium  we  must  have 

1.8  -^ —  moles  of  the  mixture  of  gases  (i.e.,  ((i  —  a)  +  2a)n, 
where  n  is  the  number  of  moles  present),  in  the  final 


THE  GASEOUS  STATE.  33 

volume  1.8  (—    —  22.4—  --  —  )  liters,  the  total  pressure 
\28o.3  273      / 

remaining   constant.     The   partial   pressures,    then,    are 

1-0.8  0.8  0.8 

ppch  =     I  g    i  PPCI,  =—g  and  />chiorine  =  :j-g  atmospheres. 

As  there  are  i  -0.8  moles  of  PC15,  0.8  moles  of  PC13,  and 
0.8   moles   of  chlorine  the    final  concentrations    (moles 

per   liter)   are  -   for   PC15,    and 


. 
28o.3  273 

0.8 


*(; 


10          273  +  250^ 
22.4 


each  for  PCla  and  chlorine. 


^280.3  273      / 

Since  in  this  case  2  moles  are  formed  from  each 
original  mole  decomposed,  the  volume  of  the  substance 
on  the  right  is  greater  than  the  volume  of  that  on  the 
left,  and  the  reaction  must  go  backward  (i.e.,  toward 
the  left),  when  the  external  pressure  is  increased.  This 
law  is  general,  and  is  expressed  by  Le  Chatelier  as  fol- 
lows: "Any  change  in  the  factors  of  equilibrium  from 
the  exterior  is  followed  by  a  reverse  change  within  the 
system."  In  other  words,  pressure  in  the  above  case 
favors  the  formation  of  the  substance  with  the  smaller 
volume.  (That  a  dissociating  system  is  an  equilibrium 
is  shown  by  the  fact  that  the  direction  of  the  reaction  is 
dependent  upon  the  direction  of  the  temperature-change). 
Such  a  reaction  as  2HI  =  H2+l2,  on  the  other  hand, 

'*'     U-IMT 


34  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

where  the  same  number  of  moles  exist  on  either  side 
(i.e.,  a  reaction  unaccompanied  by  any  volume-change) 
will  be  uninfluenced  by  a  change  in  pressure,  as  indeed 
is  observed  by  experiment. 

ii.  Variation  from  the  gas  laws.  The  equation  of 
Van  der  Waals. — The  gas  laws  are  simply  limiting  laws, 
and  while  they  hold  in  general  for  pressures  up  to  2 
atmospheres,  above  this  value  differences  are  observed. 
In  the  case  of  hydrogen  the  product  pv  is  always  higher 
than  it  should  be.  This  was  discovered  by  Natterer  and 
ascribed  by  Budde  to  the  fact  that  the  molecules  them- 
selves occupy  some  of  the  volume.  If  this  correction 
for  the  volume  is  by  then,  instead  of  (9),  we  have 

p(V-b)-RT, 

where  b  is  a  constant  for  each  gas. 
From  the  formula 

RT 
b  = --  +  V 

P 

Budde  calculated  the  value  of  b  and  found  it  to  be  equal 
to  0.00082,  at  which  point  it  remained  constant,  for 
pressures  varying  from  1000  to  2800  meters  of  mercury. 
In  the  case  of  hydrogen  this  volume-correction  is  satis- 
factory; for  other  gases,  however,  the  variation  is  some- 
what different.  In  general  for  these  the  compressibility 
at  low  pressure  is  much  greater  than  is  accounted  for 


THE  GASEOUS  STATE.  35 

by  the  Boyle-Mariotte  law,  reaches  a  minimum  and 
then  increases  so  that  the  product  pV  passes  through  the 
value  given  by  the  law.  This  shows  that  there  is  some 
factor  in  the  behavior  of  these  gases  which  is  absent 
or  negligibly  small  in  the  case  of  hydrogen.  Van  der 
Waals  ascribed  this  to  the  mutual  attraction  of  the 
molecules,  which  acts  in  the  same  direction  as  the  pres- 
sure, making  that  larger  than  it  seems  to  be.-  If  p  is 
the  pressure  and  a  the  specific  attraction,  then 


(10) 


which  is  the  equation  of  Van  der  Waals.  Here  a  is  the 
molecular  attraction  when  i  mole  occupies  the  volume 
of  i  cc.,  and  b  is  the  volume  occupied  by  the  molecules 
themselves,  p  and  a  are  expressed  most  conveniently  in 
terms  based  on  po  =  i,  where  this  is  the  original  pressure 
in  atmospheres  when  the  gas  is  confined  in  the  volume  VQ. 
V  and  b  are  then  expressed  in  terms  of  VQ. 

This  law,  although  deduced  by  aid  of  hypothetical 
assumptions,  will  hold  even  in  the  face  of  evidence 
showing  these  assumptions  to  be  false.  For  the  terms 

b    and    a    are    determined    by    experiment   (b   and    T^ 

expressing  the  discrepancy  between  the  observed  facts 
and  the  conclusions  from  the  simple  gas  laws)  ;  and  the 
worst  that  can  befall  them  is  a  loss  of  name. 


36  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

By  this  law  we  can  understand  just  how  gases  vary 
from  Boyle's  law.     For  small  pressure  and  large  volume 

the  term  ^  and  b  disappear  in  contrast  to  p  and  F,  and 

the  gas  follows  the  simple  law.  The  effect  of  the  value 
of  a  and  b  upon  the  product  pV  is  shown  by  (10)  in  the 
form 


or,  for  constant  temperature, 

a     ab 


When  V  is  large  and  p  is  small  ^  and  bp  disappear  as 

compared   with  -~;    when   —  =  (~^  +  bp\   we   have   the 

minimum  of  the  variation  of  pV  and  Boyle's  law  holds. 

These  constants  have  been  determined  for  the  dif- 
ferent gases  from  the  curves  for  pV}  and  a  few  of  the 
values  found  are  given  below: 


002=0.00874 
802  =  0.039 
Air =0.03  7 
H  =  0.0000 


THE  GASEOUS  STATE. 


37 


Air  =  0.0026 

002=0.0023 

£["=0.0067 

The  extent  of  the  variation  in  the  value  of  pV  at  con- 
stant temperature  is  shown  in  Figure  i.    If  pV  is  con- 


A0  60  80  W  120  W  160  180  SOO  220  3M  f60  W  300 

FIG.  i. — NITROGEN. 


stant,  the  curve  would  be  a  straight  line  parallel  to  the 
/>-axis  (pV  being  the  axis  of  ordinates).  The  higher 
the  temperature,  the  more  the  gas  behaves  like  hydrogen, 

The  effect  of  the  term  7^  is  shown  by  the  difference 

between  the  curve  for  N  and  for  H. 
This  difference  can  perhaps  be  shown  in  a  more  striking 


3&  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

way  by  a  comparison  of  the  'numerical  values^  of  pV. 
The  following  table  is  for  N  at  22°,  the  values  having 
been  found  by  Amagat.  p  is  given  in  atmospheres. 


p 

pv 

p 

PV 

I. 

I  .0000 

109.17 

0.9940 

27.29 

0.9894 

126.90 

1.0015 

62.03 

0.9858 

251-13 

1.0815 

80.58 

0.9875 

430.77 

I  .  2966 

We  see  by  this  that  up  to  about  12  atmospheres  pV 
decreases  in  value,  then  follows  Boyle's  law,  and  then 
varies  just  like  hydrogen.  When  a  and  b  have  been 
once  determined  for  any  pressure  they  may  be  used  for 
all  pressures.  For  ethylene  at  20°  by  Van  der  Waals' 
equation,  we  have 


Obs. 

Calc. 

31-58 

914 

%5 

84.16 

399 

392 

110.47 

454 

456 

282.21 

941 

940 

398-71 

1248 

"54 

where  p  is   in    atmospheres,  pV  is   multiplied   by  1000, 
a  =  0.00786,  and  £  =  0.0024. 

12.  Specific  heat.  The  first  principle  of  thermo- 
dynamics.— When  heat  energy  is  applied  to  a  body 
the  temperature  rises.  The  ratio  of  the  amount  of  heat 
supplied  to  the  consequent  rise  in  temperature  is  called 

the  capacity  of  the  body  for  heat,  i.e.,  •—.     The  value  of 


THE  GASEOUS  STATE.  39 

this  term  depends  naturally  upon  the  original  temperature 
of  the  body,  its  pressure,  etc. 

The  s-pecific  heat  of  any  substance  is  the  capacity  jor 
heat  of  the  unit  of  mass,  i.e., 


_ 
m   dt' 


It  has  been  observed  that  the  specific  heat  of  a  gas 
depends  upon  the  conditions  under  which  it  is  determined. 
If  the  gas  is  allowed  to  expand  under  its  previous  pressure 
the  specific  heat,  CPI  is  different  from  that  obtained  when 
the  pressure  .  varies  and  the  volume  remains  constant, 
cv.  Before  considering  the  reasons  for  this,  and  finding 
the  relation  between  the  two  values,  it  will  be  necessary 
for  us  to  inquire  into  the  nature  of  heat  energy  and  its 
possible  transformations. 

Mayer  in  1841  was  the  first  to  develop  the  subject  of 
the  theory  of  heat  and  its  transformations,  or  Thermo- 
dynamics, .as  we  know  it  to-day.  Before  that  date 
heat  was  considered  as  an  actual  substance,  which  could 
be  made  to  enter  or  leave  a  body.  Mayer  first  recognized 
it  as  an  energy,  which  could  be  obtained  from  any  other 
energy  or  turned  into  the  latter;  his  principal  work  was 
to  determine  the  equivalent  of  heat  energy,  i.e.,  a  factor 
by  which  heat  energy  can  be  given  in  units  of  mechanical 
energy.  He  was  led  to  this  conclusion  by  the  point 
already  mentioned,  that  the  specific  heat  of  gas  for  con- 


40  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

stant  pressure  is  always  larger  than  the  specific  heat 
of  the  same  for  constant  volume.  His  reasoning  was  as 
follows:  &ince  the  specific  heat  at  constant  pressure  is 
always  greater  than  that  for  constant  volume,  there 
must  be  some  condition  in  the  former  state  which  absorbs 
this  extra  heat.  In  the  case  of  constant  pressure  the 
volume  increases,  and  work  must  be  done  to  overcome 
the  atmospheric  pressure,  which  would  keep  the  volume 
constant.  It  is  a  logical  consequence,  then,  to  consider 
that  this  extra  heat  is  simply  the  amount  necessary  to 
do  the  mechanical  work  of  expansion.  In  other  words, 
a  certain  number  of  calories  are  found  to  be  equal  to  a 
definite  number  of  mechanical  units;  this  value  for  one 
calorie  is  the  mechanical  equivalent  oj  heat.  This  term 
may  be  calculated  as  follows:  The  difference  between 
the  heats  necessary  to  raise  the  temperature  of  i  gram 
of  air  i°  C.  under  the  two  conditions  is  by  experiment 

cp  —  cv  =  o.o6g2  cal. 

This  0.0692  cal.  is  the  heat  which  is  equivalent  to  the 
work  necessary  to  expand  i  gram  of  air  1/273  °f  its 
volume  at  o°.  Imagine  i  gram  of  air  at  o°  enclosed 
in  a  tube  with  a  cross-section  of  i  square  centimeter.  It 
will  occupy  the  space  of  773.3  cms.,  if  the  pressure  is 
76  cms.  oijpflg,  for  i  gram  of  air  occupies  under  these 
conditions  773-3\cc.,  which  is  the  specific  volume  of 
air.  An  increase  of  temperature  of  i°  C.  will  expand 


THE   GASEOUS  STATE.  \  41 

this  volume  1/273,  *-e->  2-^3  cms-  The  weight  of  the 
atmosphere,  1033  grams,  will  then  be  raised  >through 
this  distance.  The  work  necessary  to  do  this  i. 

1033  X  2.83  =  2923.4  gr.-cms., 

which  is  equivalent  to  0.0692  calorie.  For  one  calorie, 
then,  we  have 

2923.4 

=  42245  gr.-cms., 


which  is  the  mechanical  equivalent  oj  heat. 

Joule  transformed  a  known  amount  of  mechanical 
work  into  heat  by  friction,  and  found  from  the  heat 
developed  that 

i  cal.  =42355  gr.-cms.* 

The  great  consequence  .  of  Mayer's  work  is  what  is 
known  as  the  first  principle  of  thermodynamics.  Accord- 
ing to  this  the  energy  of  any  isolated  system  is  constant, 
i.e.,  when  energy  seems  to  disappear  it  is  simply  trans- 
formed into  another  form.  If,  for  example,  a  gaseous 
body  is  heated,  its  internal  energy  is  increased;  if  the 

*  A  liter-atmosphere  is  the  work  necessary  to  force  the  weight  of 
the  atmosphere  through  a  liter  of  space.  We  have  then 


loX  100 

i  L.  A.  =  —  '2^—  —  =24.2?  cals., 

42600 

42600  being  the  mechanical  equivalent    now    used,    the  average  of  a 
number  of  late  determinations. 

R,  the  molecular  gas  constant,  which  is  equal   to  84800  gr.-cms.,  or 
0.0821  L.  A.,  is  equal  to  2  cals. 


4  2  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

volume  increases,  however,  a  certain  amount  of  this  heat 
is  used  to  overcome  the  atmospheric  pressure,  and  the 
increase  of  the  internal  energy  is  less  than  would  other- 
wise be  the  case.  If  U  is  the  internal  energy  and  the 
amount  of  heat  dQ  is  supplied  to  the  body,  then 

dQ 


where  dW  is  the  amount  of  work  done  by  the  body  by 
virtue  of  the  heat  absorbed,  and  dU  is  the  corresponding 
increase  in  the  internal  energy.  If  we  consider  the  work 
as  overcoming  a  resistance  it  is  more  readily  handled. 
Suppose  the  gas  to  increase  its  volume,  the  pressure 
remaining  constant;  then  for  i  mole 

dW=pdV, 

where  p  is  the  intensity  factgr  and  V  the  capacity  factor 
of  volume  energy.  Substituting  this  value  of  dW  in 
the  former  equation,  we  have 


(u)  dQ 

where  dU,  dQ,  and  pdV  are  expressed  in  absolute  units. 
If  dU  and  pdV  are  given  in  mechanical  units,  then  (n) 
becomes 

dU+pdV 


where  A  is  the  mechanical  equivalent  of  heat,  i.e.,  the 
value  of  i  calorie.     We  shall  use  the  form  (n),  however, 


THE  GASEOUS  STATE.  43 

always  remembering  that  dQ,  dU,  and  pdV  are  expressed 
in  the  same  units.  The  internal  energy  U  of  a  gas  may 
be  considered  as  a  function  of  pressure  and  volume,  of 
pressure  and  temperature,  or  of  volume  and  temperature. 
Of  these  we  shall  only  consider  the  case  where  temperature 
and  volume  are  variable.  We  have  then 


dU 
The  term  ~T<7~dV,  however,  is  equal   to  zero,  for  the 

internal  energy  of  a  gas,  according  to  Gay-Lussac's 
experiment,  remains  the  same  after  a  change  in  volume 
provided  no  external  work  is  done.  By  this  experiment 
it  was  shown  that  the  expansion  of  a  gas  into  a  vacuum 
does  not  change  the  temperature  of  the  system ;  although 
there  is  a  slight  increase  in  the  temperature  of  the  exhausted 
vessel  and  a  slight  decrease  in  the  other,  they  compensate 
one  another.  Equation  (u)  becomes  then 


If  V  is  constant,  i.e.,  dV  =  o,  then 


44  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

This  term,  -T~T,  the    increase  in  the  internal  energy 

caused  by  an  increase  of  temperature,  is,  however,  the 
molecular  specific  heat  for  constant  volume;  hence 

(for  constant  volume). 


If  the  pressure  remains  constant  and  the  volume  varies, 
i.e.,  if  dV  does  not  equal  zero,  then  pdV  does  not  disap- 
pear, and  we  have 

(12)  dQ 

or 


dT~v     dT' 

This  term,  -7=,,  under  these  conditions  is  the  specific 
heat  at  constant  pressure,  i.e., 


By  differentiating  (9),  p  remaining  constant,  we  have 

pdV=RdT, 
or 

dV 


THE  GASEOUS  STATE.  45 

Substituting  this  in  (13),  we  find 
(14)  CP  =  CV+R, 

when  CP  and  Cv  refer  to  the  specific  heats  for   i   mole 
of  gas,  and  R  is  the  molecular  gas  constant. 

If  in  the  system  no  heat  is  absorbed  or  given  up  by 
conduction  or  radiation,  .i.e.,  dQ=o,  we  have  an  adia- 
batic  process.  From  (9),  by  complete  differentiation, 
we  have 

pdV+Vdp  =  RdT. 

If  in  (12)  we  eliminate  dT,  by  this  equation,  remembering 
that 


we  obtain 

(15)  dQ= 

Since  in  an  adiabatic  process  dQ=o, 


=j 

And,  representing  the  ratio  of  the  specific  heats,  T=T-,  by 
k  (for  air  =  1.41),  we  have 

Vdp+kpdV-o, 


46  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

or 


And  by  integrating  between  the  limits  p,  V,  and  p\9  FI, 
we  have 

log,  p  +  k  log,  V  =  log,  p1  +  k  log,  Fi, 
i.e., 


=   ^_Fl  -' 

o 

or,  in  another  form, 


From  (17)  we  see  that  for  an  adiabatic  process  the 
volume  changes  less  for  a  change  in  pressure,  or  the 
pressure  changes  more  for  a  change  in  volume,  than  is 
the  case  when  the  temperature  remains  constant  (i.e., 
an  isothermal  process).  For  rapid  compressions  or 
expansions,  then,  i.e.,  where  no  equalization  of  tempera- 
ture is  possible, 

#!:#::  7*:  Fi*. 

If  the  process  is  very  slow  and  the  temperature  remains 
constant,  then  we  have  (Boyle's  law) 


UNIVERSITY 
THE  GASEOUS 


+ 

Boyle's  law  holds  also  of  course  for  an  adiabatic 
change  after  the  temperature  produced  by  the  change  of 
volume  has  been  reduced  to  the  original  one. 

In  equation  (17)  we  know  that 


and 

piV 
hence 

pV       T_ 
piV^Ti 

But,  by  (17) 

V 


Whence,  by  substitution, 

£ 

V  p 

By  substituting  the  value  of  77-  instead  of  that  of  —  ,  we 

obtain 


The  temperatures  to  be  obtained  by  the  expansion 
of  a  gas  at  the  temperature  T  can  be  found  from(i8). 
Examples  of  this  are  given  in  Table  I. 


48  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

TABLE  I. 

TEMPERATURES  CAUSED  BY  ADIABATIC  EXPANSION  OF  AIR.       £=1.4. 

p  (atmospheres).       t  (initial  from  T).  I'  (final  from  TI). 

ioo  71°. 5  -201°.  5 

200  58°. 5  -213°.  5 

300  520. O  —  221°. O 

400  47°. 9  -225°. o 

500  44°.  8  -228°.  2 

The  temperatures  are  those  of  the  air  as  it  expands. 
Owing  to  the  specific  heat  of  the  vessels  used,  it  is  im- 
possible to  reach  these  temperatures  in  anything  in 
contact  with  the  gas,  although  very  low  temperatures 
may  be  thus  obtained. 

Equation  (14),  i.e., 

CP-C,=R, 


shows  that  for  all  gases  the  difference  between  the  molecu- 
lar specific  heats  is  R  =  2  cals.  This  equation  can  also 
be  derived  in  the  following  simple  way.  The  difference 
between  the  molecular  heat  at  constant  pressure  and 
that  at  constant  volume  is  obviously  equal  to  the  heat 
equivalent  of  the  work  of  expansion.  Since  the  volume- 
increase  resulting  from  a  temperature-increase  of  i°  C. 

is  1/273  times  the  volume  at  o°  C.,  i.e., ,  according 

/  \j 

to  the  notation  used  above,  and  the  amount  of  the  in- 
crease is  dependent  only  upon  the  temperature  interval 
and  not  upon  the  actual  temperature,  the  work  consumed 


THE  GASEOUS  STATE. 


49 


in  the  expansion  produced  by  an  increase  of  i°  is  . 

But  ~  -=R  (page  12),  hence 

McP-Mcv  =  Cp-C  =R, 

where  R  is  expressed  in  calories.  Naturally  the  terms 
Cp  and  Cv  are  to  be  determined  at  the  same  temperature. 
That  this  relation  holds  is  shown  by  the  following  list 
of  experimentally  determined  values  (Clausius-Ostwald). 


MOLECULAR  SPECIFIC  HEATS. 


Name. 


Const.  Pres.     Const.  Vol.     Ratio  (fc). 


Oxygen 6.96 

Nitrogen 6 . 93 

Hydrogen 6 . 82 

Chlorine 8 . 58 

Bromine 8 . 88 

Nitric  oxide  (NO) 6.95 

Carbon  monoxide ,   6.86 

Hydrochloric  acid 6. 68 

Carbon  dioxide 9.55 

Nitrous  oxide  (NzO) 9-94 

Water 8.65 

Sulphur  dioxide 9. 88 


4.96 

.40 

4.83 

•41 

4.82 

.41 

6.58 

•30 

6.88 

.29 

4-95 

.40 

4.86 

.41 

4.68 

•43 

7-55 

.26 

7-95 

•25 

6.65 

.28 

7.88 

.25 

The   ratio   k  ( =  — J  is   used   to  find   the    number  of 

combining  weights  contained  in  a  formula  weight  of  a 
gas.  This  is  purely  an  empirical  relation,  it  having  been 
observed  that  known  monotomic  gases,  mercury  for 
example,  give  £  =  1.667,  diatomic  ones  about  1.4  etc.;  the 


50  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

greater   the    number   of  atomic  weights   to  the  formula 


weight,   the  smaller  the  value    of   the  relation  — .     An 


example  of  this  application  is  given  by  Ramsay  in  the 
work  on  argon,  where  the  combining  weight,  owing  to 
the  non-existence  of  any  compounds  with  this,  is  un- 
known. 

13.  Determination  of  the  specific  heat  of  gases. — 
The  specific  heat  for  constant  volume  cv  cannot  be  de- 
termined directly  with  accuracy,  since  the  vessel  con- 
taining the  gas  absorbs  so  much  more  heat  than  the 
gas  itself.  Even  under  the  most  favorable  conditions 
the  ratio  of  the  absorption  of  heat  by  the  gas  to  that  by 
the  vessel  is  1 155. 

From  the  specific  heat  for  constant  pressure  and  the 

ratio  k——y  cv  can,  however,  be  found  indirectly. 

Cv 

The  specific  heat  at  constant  pressure  is  determined 
by  passin'g  a  certain  volume  of  the  gas,  heated  under 
constant  pressure,  to  a  certain  temperature,  through  the 
worm  of  a  calorimeter  and  observing  the  consequent 
increase  in  the  temperature  of  the  water.  We  krfow 
then  the  number  of  calories  which  causes  a  certain  tern-  < 
perature  to  exist  in  the  known  volume  of  the  gas;  and 
from  this  data  it  is  easy  to  calculate  the  specific  heat,  cp. 
The  value  of  the  molecular  specific  heat  for  constant 
pressure  for  all  gases  seems  to  converge  toward  the  value 


THE  GASEOUS  ST^TE.  St 

6.5  at  the  absolute  zero,     We  have  in  general  the  relation 


where  a  is  a  constant  for  each  gas  and  is  the  larger  the 
more  complex  the  formula.  For  a  we  find  the  following 
Values:  for  H2,  N2,  O2,  and  CO,  a  =©-.001,  NH$=  010071, 
CO2^=  0.0084,  C6H6  =  o.o5io,  ether  =  0.0738,  H2O  = 
0.0089,  C2H4  =0.0137,  CHCla  =  0.0305,  C2H5Br  =  0.0324, 
CH3COOC2H5  =  0.0674,  C3H6O  =0.0403,  N2O  =0.008^ 
This  formula  may  be  used  in  the  following  way  to  find 
the  value  cp.  For  CeHe,  a  =0.0510;  hence  at  50°,  since 
CM  =  C  we  have 


6.5+0.0510X323 

--          =0.295, 


while  Wiedemann  found  0.299  experimentally. 

The  ratio  £  =  —  can  be  found  easily  by  the  method 
cv 

of  Clement  and  Desormes. 

A  glass  balloon,  holding  about  20  liters,  is  provided 
with  a  brass  stop-cock  and  a  manometer.  From  this 
vessel  the  air  is  partially  rarefied  and  the  pressure  ob- 
served by  aid  of  the  manometer.  Let  this  initial  pres- 
sure be  pQ  and  the  atmospheric  pressure  be  P.  If  the 
cock  is  now  opened  for  half  a  second  air  will  rush  in 
until  the  external  and  internal  pressures  are  the  same. 
As  the  air  goes  in,  however,  heat  will  be  developed, 


52  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

which,  as  it  is  not  removed,  will  increase  the  tempera- 
ture of  the  gas  according  to  the  law  of  adiabatic  com- 
pression (p.  46).  By  the  process  we  have  increased 
the  pressure  from  pQ  to  P.  If  the  initial  specific  volume 
is  v0  and  the  final  one  is  v,  then  k  can  be  determined 
from  the  equation 


The  final  specific  volume  is  not  known  yet,  however. 
To  find  this  we  wait  until  the  flask  and  air  have  been 
reduced  to  the  temperature  of  the  surrounding  air  and 
again  measure  the  pressure.  If  this  pressure  is  p  at  the 
temperature  /,  then 


or 


po     \po/ 
and 

log  P-log 


log  ^-log^o' 

In  one   experiment   with   air  P  =  1.0036,   #o=o-9953> 
and  p  =1.0088  atmospheres;  hence 

£  =  1.3524. 

14.  The    second    principle   of    thermodynamics. — By 

the  first  principle  we  have  found  that  a  certain  amount 


THE  GASEOUS  STATE.  53 

of  heat  is  equivalent  to  a  certain  amount  of  work.  If 
we  consider,  however,  the  possible  means  of  transform- 
ing heat  into  work  we  find  that  a  certain  condition  must 
be  fulfilled,  i.e.,  the  heat  must  go  from  a  warmer  to  a 
colder  body.  In  other  words,  heat  can  only  produce 
work  by  going  from  a  higher  to  a  lower  temperature. 
This  condition  has  been  expressed  in  other  words.  For 
example,  heat  of  itself  can  never  go  from  a  colder  to  a 
warmer  body  without  work  being  used  upon  it.  A  per- 
petual motion  oj  the  second  kind  is  impossible. 

A  device  to  produce  a  perpetual  motion  of  the  second 
kind  would  be  a  machine  which  would  run,  for  example, 
from  the  heat  of  the  sea.  It  is  related  to  the  second  just 
as  an  ordinary  perpetual  motion  is  related  to  the  first 
principle. 

Since  heat  can  only  be  transformed  into  work  by  a 
transference  of  heat  from  a  higher  to  a  lower  tempera- 
ture, it  is  important  to  know  the  relation  which  exists 
between  the  amount  of  heat  transferred  as  heat  and 
that  transformed  into  work. 

For  this  investigation  it  is  necessary  that  all  the  work 
done  be  external  work,  which  can  be  readily  observed, 
and  then  compared  with  the  known  amount  of  heat. 
We  employ  for  this  purpose  a  process  which  was  origi- 
nated in  1824  by  Sadi  Carnot  and  which  is  known  as 
the  Carnot  cycle.  The  arrangement  is  such  that  the 
final  state  is  identical  with  the  initial  one,  i.e.,  before 


54  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

and  after  the  operation  the  body  contains  exactly  the 
same  amount  of  energy.  Then  the  relation  between  the 
heat  transferred  and  the  work  done  is  very  readily  ob- 
tained. Naturally  no  heat  must  be  lost  by  radiation  or 
conduction,  or  the  final  result  will  be  incorrect;  for 
this  reason  the  process  is  an  ideal  one  and  cannot  be 
realized  practically.  Since  all  transference  of  heat  and 
work  is  assumed  to  take  place  without  differences  in 
temperature  and  pressure  the  process  may  go  in  either 
direction,  i.e.,  it  is  reversible.  This  condition  of  rever- 
sibility is  never  obtained  in  practice,  so  that  the  relation 
which  we  find  is  the  limit  under  the  most  favorable 
conditions. 

15.  The  cycle.  Entropy. — We  assume  the  process  to 
take  place  in  four  steps: 

i.  Assume  an  ideal  gas  enclosed  in  a  cylinder  with  a 
movable  piston  at  a  certain  temperature  and  pressure. 
The  cylinder  is  now  placed  in  a  heating-bath  at  the 
temperature  7\,  and  the  volume  is  allowed  to  increase 
under  a  constant  pressure  which  is  just  greater  than 
that  of  the  atmosphere.  By  this  expansion  the  gas  will 
cool.  Here,  however,  we  .assume  heat  to  be  absorbed 
by  it  to  such  an  extent  that  the  temperature  remains 
constant.  If  the  heat  absorbed  by  the  gas  is  Qit  its 
initial  volume  Vi,  and  its  final  one  v2,  and  the  constant 
temperature  TI,  then  the  work  done  by  the  gas  will  be 


THE   GASEOUS  STATE.  55 

equal  to  jf  2pdv.      Since  the   temperature   remains  con- 
stant, equation  (12)  becomes 


rT 

or,  since  p=~  , 


or 

(20)  Gi-rJ,  lo&S 

2.  The  gas  is  next  allowed  to  expand  adiabatically 
until  the  temperature  falls  to  T2.  For  this,  the  new 
volume  being  v%,  we  have  the  relation 


3.  Next  the  pressure  is  increased  until  the  volume 
decreases  to  v4,  heat  being  removed  to  the  amount  Q2y 
so  that  the  temperature  remains  constant  at  T2.  The 

work  done  here  by  the  gas  is—  J  *pdv,  we  have  then 


(22) 


4.  Finally,  the  gas  is  compressed  adiabatically  until 
the  original  volume  v\  and  the  original  temperature  T\ 


56  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

are  reached.     This  can  be  arranged  by  choosing  v±  of 
the  proper  size.     We  have  for  this  relation 


I*JV-1\ 

TI  W 


We  have  thus  carried  the  gas  through  a  series  of  changes, 
and  have  finally  the  same  state  as  that  from  which  we 
started.  The  amount  of  heat  Qi  is  absorbed  at  the 
higher  temperature  TI,  and  a  smaller  amount  Q%  is  given 
up  at  the  lower  temperature  T^  and  a  certain  amount 
has  been  transformed  into  work.  The  amount  of  heat 
which  is  equivalent  to  the  work  done  is  equal  to  Q  =  Qi 
-Q2.  The  heat  Q2  has  simply  been  transferred  from  the 
temperature  TI  to  T2. 

The  relation  between  Qi  and  Q2  is  given  by  equations 
(20)  and  (22) 


By  (21)  and  (23),  however, 


. 
-=-,    Le,    log,-=log,-; 


hence 


THE   GASEOUS  STATE.  57 

i.e.,  the  amounts  of  heat  absorbed  and  liberated  are  pro- 
portional to  the  absolute  temperatures. 

The   amount   of  heat  transformed  into  work  is  Q  = 
(?i~ (?2-     We  have  then 

Qi-Q2Tl-T2 


and 


The  heat  transformed  into  work  by  any  reversible  process 
is  to  that  transferred  from  the  higher  to  the  lower  tem- 
perature as  the  difference  in  temperature  is  to  the  lower 
absolute  temperature.  Or  from  the  other  standpoint, 
The  work  in  calories  necessary  to  transfer  a  certain  amount 
of  heat  from  one  temperature  to  a  higher  one  by  a  reversible 
process  is  to  the  amount  of  heat  as  the  temperature  interval 
is  to  the  final  high  absolute  temperature. 

For   T2  =  o°,  Q2  will  equal  o  and  Q\  —  Q2  =  Qi,  i.e., 
at  the  absolute  zero  all  heat  is  transformed  into  work.     For 

Ti  —  T2. 

all  higher  temperatures  — ^ —  is  smaller  than  i. 

1  2 

If  the  difference  in  temperature  T\  —  T2  is  very  small 
we  can  substitute  for  it  dT,  and  as  the  difference  between 


5§  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

the  amounts  of  heat  will  then  also  be  small,  we  may 
substitute  dQ  for  Qi—  Qz'  we  find  then 

dQ_dT 
Q~  T' 

If  now  we  consider  the  heat  liberated  as  negative  and 
that  absorbed  as  positive,  i.e.,  Q2  negative  and  Q\  positive, 
then 

1,62 


or 


or 


where  there  are  as  many  terms  as  there  are  tempera- 
tures. Q  is  the  amount  of  heat  absorbed  or  emitted. 
Changing  the  sign  of  summation  to  that  of  integration, 
and  we  have 


/dO 
T=o. 


This  is  the  analytical  expression  of  the  second  princi- 
ple for  reversible  processes.  In  words  ft  means  that  for 
any  REVERSIBLE  process  the  transformation  oj  heat  into 
work  takes  place  in  such  a  way  that  the  sum  of  the  amounts 


THE  GASEOUS  STATE.  59 

of  heat  absorbed  and  liberated,  each  divided  by  its  corre- 
sponding temperature,  is  equal  to  zero. 

If  T  remains  constant  dQ=o,  i.e.,  just  as  much  heat 
is  liberated  as  is  absorbed,  and  no  heat  is  transformed 

into  work. 

/j/~\ 
-~,    found    as    above,    was    called    by 

Clausius,  the  entropy.     If  this  term  is  represented  by 
s,  then 


dQ 


When  dQ  =  o,  ds  =  o,  i.e.,  when  a  substance  by  a  re- 
versible change  neither  gives  up  nor  absorbs  heat  its 
entropy  remains  constant.  The  two  terms  vary  to- 
gether and  have  the  same  signs.  For  this  reason 
adiabatic  changes  are  also  called  isentropic.  T  always 
decreases  in  any  change,  for  heat  can  only  go  from  a 

dQ 
warmer  to  a  colder  body.      ~r=s    therefore    increases 

during   any   change,   i.e.,    the   entropy   tends   toward   a 
maximum. 

16.  The  factors  of  heat  energy. — We  have  already 
found  that  the  temperature  is  the  intensity  factor  of 
heat  energy.  The  capacity  factor,  entropy,  is  the  ratio 
of  the  heat  absorbed  or  liberated  at  constant  temperature 


60  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

to  the  absolute  temperature  of  the  process.  Therefore 
entropy  times  temperature  is  equal  to  heat  energy  0  We 
have  in  general,  then,  for  all  chemical  processes 

?-*;. 

dQ-Tds. 


CHAPTER  in. 
THE   LIQUID   STATE. 

17.  Distinction  between  liquids  and  gases. — Liquids 
are  distinguished  from  gases  by  the  fact  that  they  pos- 
sess a  volume  of  their  own,  which,  although  dependent 
upon  pressure  and  temperature,  cannot  be  changed  to 
any  extent  by  them. 

From  the  standpoint  of  experiment  all  we  can  say  of  a 
liquid,  as  compared  to  a  gas,  is  that  it  contains  less  energy 
than  the  former,  for  energy  is  always  absorbed  by  a  liquid 
when  it  is  being  transformed  into  a  gas.  On  hypothety 
ical  grounds  it  is  said  that^  the  attraction  betw^fti  tne 
molecules  of  a  liquid  is  greater  than  between  those  of 
a  gas.  -  ififtcfrth  ty&tQ  Tpf  &&Wi 

Since  the  volume  in  the  gaseous  state  is  always  greater 
than  that  in  the  liquid  state,  a  greater  temperature  is 
necessary  to  cause  a  liquid  to  boil  when  the  pressure 
over  it  is  increased.  And,  conversely,  a  liquid  boils 
more  readily  (i.e.,  at  a  lower  temperature),  when  the 
pressure  over  it  is  decreased.  (See  Le  Chatelier's 

theorem,  p.  33). 

61 


62  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

18.  Connection  between  the  gaseous  and  liquid  states. 
— If  at  constant  temperature  a  gas  is  subjected  to  a  con- 
stantly increasing  pressure,  its  state  may  change  in  one  of 
two  ways  according  to  the  conditions : 

a.  This  case  has  already  been  considered  under  gases. 
The  volume  at  first  changes  more  rapidly  than  the  pres- 
sure, next  in  the  same  ratio,  and  finally  more  slowly. 
When  the  pressure  becomes  very  high  a  further  increase 
has  but  a  slight  effect  upon  the  volume. 

b.  Here    the    relation    between   pressure    and    volume 
is  quite  different,  although  the  first  step  is  the  same, 
i.e.,  the  volume  changes  more  rapidly  than  the  pressure. 
The  ratio  here,  however,  in  contrast  to  a,  increases  con- 
tinually.    When   a   certain   pressure   is   reached   a   new 
phenomenon  is  observed:    the  gas  is  no  longer  homo- 
geneous;   one  part  has  separated  which  behaves  differ- 
ently from  the  rest,  i.e.,  the  gas  is  partly  liquefied.     For 
a  constant  temperature  this  liquefying  pressure  remains 
constant,   while  the  volume  decreases  steadily,   i.e.,   for 
one  pressure  we  have  a  whole  series  of  volumes.     An  in- 
crease in  the  external  pressure  has  no  lasting  effect  upon 
the   internal   pressure    (which   still   remains   as   it   was), 
but  it  causes  the  volume  to  decrease  more  rapidly.     Only 
after  the  whole  gas  has  been  liquefied  can  the  pressure 
be  increased.     Then,  however,  we  shall  compress  only 
the  liquid,  just  as  in  the  last  stage  of  a  we  compressed  thq 
gas. 


THE  LIQUID  STATE.  63 

The  condition  which  causes  a  gas  under  compression 
to  follow  a  or  b  is  the  temperature.  If  this  is  above 
a  certain  point,  which  depends  upon  the  nature  of  the 
gas,  process  a  will  be  followed ;  if  below  this  point,  process 
b.  This  was  the  first  recognized  by  Andrews  in  1871.  If, 
for  example,  we  compress  CC>2  gas,  keeping  the  tem- 
perature at  o°,  the  volume  changes  more  rapidly  than  the 
pressure,  and  at  a  pressure  of  35.4  atmospheres  the  gas 
condenses  to  a  liquid.  The  higher  the  temperature, 
under  3o°.92,  the  higher  the  pressure  must  be  to  cause 
condensation;  but  above  3o0.Q2  no  condensation  is 
possible.  Thus: 


t  p 

31°,  impossible  to  liquefy  it. 
30°  .92  =  73.6  atmospheres. 

30°=  73-o 
130.1  =  48.9 


t       p 

—  2i°=2i.5  atmospheres. 

-4o°=n.o 
-590-4  =  4-6 
-7o°.6  =  2.3 

-78°=    1.2 


Andrews  called  the  temperature  30°. 9 2  the  critical 
temperature.  Correspondingly  we  call  the  pressure  which 
is  necessary  to  liquefy  the  gas,  at  the  critical  temperature, 
the  critical  pressure  (73.6  atmos.  at  3O°.9),  and  the  volume 
which  the  gas  or  liquid  occupies  (dt  =  dg),  under  these 
two  conditions,  the  critical  volume. 

The  best  method  of  showing  the  behavior  of  a  gas 
under  compression  is  by  plotting  a  curve  in  a  system 
of  coordinates  where  the  pressures  are  laid  out  upon  the 


ELEMENTS   OF  PHYSICAL  CHEMISTRY. 


axis  of  ordinates,  and  the  volumes  upon  the  axis  of 
abscissae.  In  these  two  curves  the  horizontal  parts, 
which  show  constant  pressure  for  varying  volume,  in- 


•4.05 

•iOO 

95 

90 

45 

SO 

•15 

-10 

•65 

-60 

-55 

-50 


FIG.  2. 

dicate  that  the  gas  liquefies.  The  vertical  parts  refer 
to  the  liquid  state,  while  all  others  refer  to  the  gaseous 
state. 

Fig.  2  shows  the  behavior  of  CO2  and  air,  the  ordi- 
nates being  pressures  in  atmospheres.  At  31°.!  CO2 
there  is  no  horizontal  part,  i.e.,  no  varying  volume  for 


THE  LIQUID  STATE.  65 

constant  pressure;  the  gas  does  not  liquefy,  since  it  is 
above  its  critical  temperature.  For  all  temperatures 
below  31°. i  the  horizontal  part  is  present,  i.e.,  the  gas 
liquefies*  under  a  sufficient  pressure.  Under  all  these 
pressures  air  remains  a  gas  and  behaves  as  CC>2  does 
when  above  3i°C. 

When  a  solution  is  heated  to  its  critical  temperature 
it  is  observed  that  there  is  no  separation  of  solvent  and 
solute,  and  that  apparently  the  gaseous  solvent  has  the 
same  power  to  dissolve  the  substance  that  the  liquid  one 
had. 

19.  Vapor-pressure  and  boiling-point. — The  vapor- 
pressure  of  a  liquid  is  the  pressure  to  which,  at  any 
fixed  temperature,  it  will  evaporate  in  an  exhausted 
space.  If  the  vapor  thus  formed  is  removed,  naturally 
more  vapor  will  form,  and  by  successive  removals  of 
portions  in  this  way  the  entire  amount  of  liquid  may 
be  evaporated.  To  cause  a  liquid  to  evaporate  contin- 
uously, then,  it  is  necessary  to  remove  the  pressure  of  the 
vapor  above  it.  This  may  be  done  as  above,  or  the  same 
condition  may  be  attained  by  making  the  exhausted  space 
large  enough  to  contain  the  vapor  of  all  the  liquid  and 
yet  not  reach  a  high  enough  pressure  to  prevent  further 
evaporation.  This  latter  condition  is  indeed  obtained 
in  the  open  air  when  the  vapor  pressure  of  the  liquid  is 
increased  to  such  an  extent  by  heat  that  it  exceeds  the 
counter-pressilre  of  the  atmosphere.  The  temperature 


66  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

necessary  for  this  is  called  the  boiling-point  of  the  liquid, 
the  boiling-point  designating  that  temperature  at  which 
liquid  and  gas  can  exist  together  in  all  proportions.  Every 
external  pressure  corresponds  to  a  certain  temperature, 
and  these  two  magnitudes  increase  and  decrease  together. 
At  the  critical  temperature  the  significance  of  vapor- 
pressure  is  naturally  lost.  For  every  pressure  under 
the  critical  one,  however,  there  is  a  certain  temperature 
at  which  gas  and  liquid  may  exist  together  in  all  propor- 
tions: this  temperature  is  the  boiling-point  at  that  pres- 
sure. Correspondingly,  for  eveVy  temperature  under 
the  critical  one,  there  is  a  certain  pressure  at  which  gas 
and  liquid  may  exist  in  equilibrium  in  all  proportions: 
this  is  the  vapor- pressure  at  that  temperature. 

Vapor-pressure  of  a  liquid  may  be  determined  in  one 
of  several  ways.  One  method  of  theoretical  interest 
which  gives  good  results  is  by  J.  Walker.  A  current  of 
dry  air  is  passed  through  the  liquid  which  is  held  at 
constant  temperature  and  the  loss  in  weight  of  the  liquid 
observed.  The  air  in  passing  through  the  liquid  will 
absorb  an  amount  of  vapor  proportional  to  the  vapor- 
pressure  of  the  liquid.  If  V  is  the  volume  of  air  which 
in  passing  through  the  liquid  absorbs  the  weight  of  I 
mole,  then  in  that  volume  we  have  the  relation 

pV  =  RT '(where  p  is  the  vapor-pressure). 
If  v  is  the  volume  of  air  which  absorbs  x  grams  of  the 


THE  LIQUID  STATE.  67 

liquid  with  the  molecular  weight  M  in  the  gaseous  state, 
then  we  have 


or 


The  term  v  here  represents  the  total  volume,  i.e.,  that 
of  vapor  and  air,  but  that  of  the  latter  may  be  taken  as 
a  rule  as  the  volume  of  the  other  is  small  in  comparison. 

If  we  compare  two  liquids  at  the  same  temperature, 
passing  equal  volumes  of  air  through  them, 

L-?L    Ml. 
p~MX  x" 

or 

x     M' 


where  M  and  M'  are  the  molecular  weights  of  the  sub- 
stances in  gaseous  form. 

This  equation  may  also  be  used  for  one  liquid  alone, 
that  is  in  the  form 


Here  naturally  we  must  assume  that  the  saturated  vapor 
behaves  as  a  more  permanent  gas  would,  i.e.,  obeys  the 


68  ELEMENTS  OP  PHYSICAL  CHEMISTRY. 

gas  laws.  The  only  difficulty  in  the  use  of  these  formulas 
is  the  necessity  o]  having  the  same  dimensions  on  either 
side.  If  p  be  expressed  in  atmospheres,  V  in  liters  and 
R  in  liter-atmospheres  little  difficulty  will  be  experienced. 

>v 

oc  and  M  are  then  found  in  grams,  the  ratio  -TT  being 

equal  to  the  number  of  moles  going  info  the  gaseous 
state.  Although  this  method  has  been  used  by  several 
investigators,  apparently  with  success,  it  has  lately  been 
the  subject  of  a  polemic  by  Carveth-Fowler  and  Perman 
(J.  Phys.  Chem.,  8  and  9,  'o4~'o5). 

20.  The  heat  of  evaporation. — For  the  transformation 
of  a  liquid  into  its  gaseous  form  a  considerable  amount 
of  heat  is  necessary.  There  are  two  causes  for  this 
absorption  of  heat.  The  volume  must  be  increased 
against  atmospheric  pressure,  and  the  liquid  must  be 
transformed  into  the  gaseous  state.  The  former  is  of 
the  least  value.  Its  amount  may  be  calculated  from 
pV  =  RT  =  2T  cals.;*  i.e.,  for  every  mole  of  water- 
vapor  formed  from  liquid  water  2  X  (2  73  +  100)  =746 
cals.  are  used  for  the  expansion.  Since  the  heat  of  evapo- 
ration of  i  mole  (18  grams)  of  H^O  to  steam  at  100°  is 
9650  cals.,  and  the  amount  necessary  for  the  expansion 


*  Where  V  represents  the  volume  of  the  gas,  which  is  so  large  that 
that  of  the  liquid  may  be  neglected.  pV  then  represents  the  actual  work 
of  expansion  against  the  constant  atmospheric  pressure,  p,  i.e.,  is  equiva- 
lent to  p(Vg-Vi). 


THE  LIQUID  STATE.  69 

is  but  746  cals.,  it  will  be  observed  that  the  difference 
in  energy  -content  of  the  two  states  is  very  considerable. 

The  ratio  of  the  molecular  heat  of  evaporation  (=  mo- 
lecular weight  multiplied  by  the  heat  of  evaporation 
of  i  gram)  to  the  absolute  boiling-point  has  been  found 
experimentally  by  Trouton  to  be  a  constant  for  a  very 
large  number  of  substances,  and  equal  to  approximately 
21,  i.e., 

Mw 


By  aid  of  this  equation  unknown  heats  of  evaporation 
may  be  calculated  (with  a  possible  error  of  5%)  from 
the  molecular  weight  and  the  boiling-point.  By  this 
formula  it  is  also  possible  to  find  the  molecular  weight 
of  a  liquid,  provided  we  know  the  latent  heat  of  evapo- 
ration. Longuinine  found  in  this  way  that  liquid  acetic 
acid  molecules  are  twice  the  size  of  the  gaseous  ones, 
i.e.,  instead  of  finding  21,  using  the  molecular  weight  as 
found  from  the  gaseous  state,  he  obtained  about  13.  If 
the  molecular  weight  used  is  doubled,  26  instead  of  21 
is  found,  showing  that  undoubtedly  an  association  does 
take  place. 

Van  Laar  explains  the  maximum  density  of  water  at 
4°  as  dependent  upon  association  which  increases  with 
decreasing  temperature.  The  liquid  itself  without 
association  would  increase  in  density;  the  association 


7o  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

causes  the  density  to  decrease.  At  4°  these  two  influence 
just  compensate  each  other,  and  the  liquid  has  its  max: 
mum  density.*  Below  this  point  the  influence  of  th 
association  is  the  stronger,  and  consequently  the  volurn 
increases. 

Young  and  Thomas  have  also  found  an  empirics 
relation  from  which  molecular  weights  of  liquids  ma 
be  determined.  It  is 

MPk 
DkTk~ 

where  M  is  the  molecular  weight  of  the  liquid,  Pk  th 
critical  pressure,  Dk  the  critical  density,  and  Tk  th 
critical  temperature  (absolute).  We  have  for  CCl± 

22XDkXTk    22X0.556X556 

M  =  -  n  -  =  -  =  Ic?Ii 
Pk  44-9 

Theoretical  =  154, 
and  for  acetic  acid 


,93>  Theoretical=6o. 


We  see  then  from  this  that  acetic  acid  is  associated  eve 
at  the  critical  state. 

*  1  8  gr.  H2O  going  into  the  state  (H2O)2  would  increase  in  volume  b 

8.44  cc.,  46  gr.  C2H6O  in  (C2H8O)2  by  20  c.,  etc.     Van  Laar  also  explair 

v    the  increase  in  volume  when  alcohol  and  water  are  mixed  to  a  mutu< 

breaking  down  of  the  associated  forms  (H2O)2  and 


THE  LIQUID  STATE.  71 

21.  The  relation  between  vapor-pressure,  heat  of 
evaporation,  and  temperature.  —  An  equation  showing 
the  relation  between  these  three  quantities  can  be  found 
as  follows:  Equation  (6)  gives  the  relation  between 
any  two  kinds  of  energy  when  in  equilibrium.  We 
have 

SdT=cdi=vdp, 

or,   since   S  can   only   be   determined   as   a   difference, 
(Si-S2)dT=(vi-v2)dp,    where    Si    is    larger   than    S2 
and  refers  to  the  state  which  absorbs  heat  when  formed 
(gaseous). 
Since  S  varies  with  the  amount  of  heat,  we  may  write 


(01       C?2\  W 

—  —  —}  we  may  substitute  -~,  where  w  is  the 

latent  heat  of  vaporization  for  i  gram  of  liquid.    We 
have  then 

w  dp  T(vi-v2)     dT 

T(v1-v2)=df  w        =~dp' 

Since  the  volume  of  gas  is  large  compared  to  that 
of  the  liquid,  we  may  neglect  the  latter,  and,  provided 
the  gas  obeys  Boyle's  law,  find  the  former  from 


pV-RT,     F=—  . 


72  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

We  have  then,  changing  w  to  /,  i.e.,  the  molecular 
latent  heat,  and  vi  to  F,  which  is  the  volume  of  i  mole 
of  the  gas, 

JL_dp_        ,1     2T2     dT 
^f~2==df  ~pT  =  Jp' 

It  is  necessary  to  remember  in  using  these  equations 
that  2  (=R)  and  /  are  in  calories  and  p  in  mechanical 

dp 
units,  -j~  gives  us  the  change  in  vapor-pressure  for  i° 

dT 
of  temperature,   and   -j—  gives   the   change   in   boiling- 

point  for  unit  of  external  pressure.  To  get  a  correct  re- 
sult, then,  care  must  be  taken  with  the  dimensions  (see 
pp.  68  and  43). 

Example.  —  The  boiling-point  of  ethyl  iodid  at  73.3 
cms.  is  72°.2  C.,  /  =  7597;  find  change  in  vapor-pressure 
for  i°  C.  We  have 


2X(345-4)2 


Another  relation  involving  the  heat  of  evaporation  has 
been  derived  by  Crompton  (Proc.  Chem.  Soc.,  17,  61, 
1901)  and  gives  very  good  results.  Assume  that  in  a 
saturated  vapor  the  equation  pV=RT  holds,  and  that  it 
is  possible  at  constant  temperature  and  by  compression 
alone  to  reduce  its  volume  V  g  to  that  occupied  by  the 
liquid  F/,  without  any  change  in  state  occurring  and 


THE  LIQUID  STATE.  73 

the  gas  continuing  to  obey  the  law  throughout  the  com- 
pression. The  work  done  in  causing  this  change  of 
volume  will  be  equal,  then,  to 


/v 
gRT 
fW- 


and  as  no  change  in  temperature  takes  place  heat  equiva- 
lent to  this  will  be  given  out.  The  gas  will  now  occupy 
the  volume  which  would  be  occupied  by  the  liquid,  but 
of  itself  it  will  not  retain  its  volume.  In  order  to  obtain 
it  in  such  a  state  that  it  will  do  this  it  is  necessary  to 
remove  its  ability  to  expand.  By  removing  an  amount 

of  energy  equal  to   that  expended   in  compressing  the 

y 
material,    viz.,    RT   log/rr,  we  shall,  however,  also  re- 

move this  power  of  expansion.  The  total  heat  given  out, 
then,  by  the  production  of  the  liquid  from  the  vapor 

y 

is  2R  T  log/TT,  and  this  is  equal  to  the  latent  heat  of 

evaporation  of  the  molecular  amount,  i.e., 


y 
or,  since  the  ratio  of  the  molecular  volumes  -*?-  under 

normal  conditions  is  the  same  as  that  of  the  specific 


74  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

volumes,    or   inversely   as    the     specific     densities,,  and 
using  Brigg's  logarithms,  we  have 


. 
-4343 


Thus,  since  for  CC>2  at  7^  =  248,  ^=1.10,  dg=  0.044, 

M  =  44,   we   find   7^  =  71.91    calories,   where   the   experi- 

mental value  is  72.23. 

Any  association  in  the  liquid  state  naturally  fails 
to  be  accounted  for  in  this  equation.  For  liquids  which 
are  shown  in  other  ways  to  be  associated  (i.e.,  apparently 
larger  in  molecular  weight  than  in  the  gaseous  state) 
the  calculated  value  is  found  to  be  too  low,  the  molecu- 
lar aggregations  on  liquefaction  apparently  involving 
the  evolution  of  heat.  This  formula,  according  to  Mills, 
gives  very  accurate  values  at  temperatures  at  which  the 
vapor-pressure  is  high,  but  values  which  are  too  large 
for  those  temperatures  giving  low  vapor-pressures;  the 
divergence  being  greater  than  one  calorie  only  for  a 
few  cases.  (J.  Phys.  Chem.,  8,  p.  593,  1904.)  The 

term  —=-   of  Trouton's  law  (p.  69)   is  equal  according 

to  this  equation  to  2R(\oge  Vg—  logeFj),  which  for  many 
liquids  gives  values  lying  between  22.8  and  23.9. 

The  heat  of  evaporation,  however,  also  varies  with 
the  temperature.  The  expression  for  this  variation  is 


THE  LIQUID  STATE.  75 

derived  as  follows  :  (i)  Allow  i  mole  of  liquid  to  vaporize 
at  the  boiling  temperature  T,  and  then  heat  this  volume, 
holding  it  constant,  to  the  temperature  T  +  t.  The  gas 
in  expanding  has  done  the  work  pV=RT  and  has  ab- 
sorbed the  heat  Cvt  in  going  to  the  higher  temperature, 
while  the  latent  heat  of  evaporation  is  IT-  (2)  We  reach 
the  same  final  state  by  first  heating  the  liquid  to  T+t, 
by  which  the  heat  Me  is  absorbed,  and  then  allow  the 
gas  to  form  at  T+t.  The  work  of  expansion  here  is 
pV=R(T  +  f),  and  the  latent  heat  lT+t.  By  these  two 
processes  we  have  reached  the  same  final  state  from  the 
same  initial  one,  so  by  the  principle  of  the  conservation 
of  energy  the  loss  in  energy  is  the  same  in  both  cases. 
We  have  then 


-(lT  +  Cj)=RT+Rt-(lT+t+Mct), 
or 


and,  by  (14), 


dl  dw 

c~c  = 


i.e.,  the  latent  heat  of  evaporation  increases  for  each 
degree  by  an  amount  equal  to  the  specific  heat  of  the 
gaseous  state  at  constant  pressure,  minus  the  specific 
heat  of  the  liquid.  Since  the  specific  heat  of  the  gas-  is 


76  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

always  less  than  that  of  the  liquid,  the  increase  is  negative, 
i.e.,  the  heat  of  evaporation  always  decreases  with  in- 
creasing temperature,  c  for  benzene  at  50°  is  0.45,  cp  for 

50°  (see  p.  51)  =0.295,  hence  ^  =  0.295-0.45=  -0.155, 

for  which  —0.158  has. been  found  by  experiment,  the 
difference  being  within  the  experimental  error. 

The  substance  must  have  the  same  molecular  weight 
in  both  the  gaseous  and  liquid  state  if  correct  results 
are  to  be  obtained  by  this  formula.  If  associated  in 
liquid  form,  the  result  here  will  be  too  small,  owing 
to  the  fact  that  Mc  =  C  will  be  smaller  than  the  proper 
C.  It  has  been  found  experimentally  and  can  be  proved 
theoretically  that  the  specific  heat  in  the  liquid  state  and 
that  in  the  gaseous  state  bear  a  constant  ratio  to  each 
other,  so  that  unless  the  molecular  weight  is  different  in 

dw 
the  two  states,  -j~  is  the  same  for  all  temperatures.     For 

ethyl  alcohol  there  seems  to  be  an  association  at  10°;  we 
have  /  =  220.9  at  °°>  2 1 1. 2  at  10°,  and  220.6  at  20°. 

22.  Refraction  of  light. — La  Place,  in  determining 
the  velocity  of  light  in  different  media,  found  the  relation 

7^2  —  j 

— -j—  =  constant  for  all  temperatures 

when  n  is  the  refraction  of  light  and  d  is  the  density  of  the 
medium.  For  many  liquids  this  holds  true,  approximately, 


THE  LIQUID  STATE.  v  77 

but  Gladstone  and  Dale  found  empirically  a  relation 
which  holds  more  exactly  and  for  many  more  liquids 
than  the  former.  This  is 


n-i 

—-j-  =  constant  for  any  one  liquid. 


Further,  in  the  case  of  a  mixture,  the  constant  is  found 
to  be  equal  to  the  sum  of  those  for  the  ingredients.  We 
have  then 

H — I  HI  —  I  W/2 —  I 


where  p  is  equal  to  100,  and  pi  and  p2  represent  the 
percentages  by  weight  of  the  two  ingredients.  By  aid 
of  this  it  is  possible  to  make  an  optical  analysis. 

Example.  —  An  unknown  mixture  of  amyl  and  ethyl 
alcohols  gives  n  =  1.3822,  ^=0.8065.  What  is  the  per- 
centage composition  of  each  in  the  mixture? 

For  amyl  alcohol  n\  =  1.4057,   di  =0.8135;    f° 
alcohol  #2  =  i  -3606,  </2  =  o.8n;  hence 

100  X  0.4738  =  pi  X  0.4987  +  p2  X  0.4501  ; 


and    since    pi  +  p2  =  ioo,    we    find    ^1=48.8,    p2  =  $i.2. 
By  direct  weighing  it  was  found  that  p\  is  48.9  and  p2 

is  51.1. 


7«  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

This  formula  also  holds  approximately  for  solutions, 
i.e.,  for  finding  the  coefficient  of  refraction  in  the  solid 
state  from  that  of  the  solution  and  the  solvent. 

23.  Surface-tension.  —  The  surface-tension  x  oj  a 
liquid  is  the  work  which  is  necessary  to  form  a  surface 


FIG.  3. 

of  one  square  centimeter  (millimeter)  in  area.  Some 
liquids  wet  the  walls  of  a  glass  tube,  while  others  do  not, 
and  upon  this  depends  the  shape  of  the  surface  formed. 
Imagine  a  plate  of  glass  suspended  vertically  in  a  vessel 
of  water,  i.e.,  a  liquid  which  wets  the  glass.  The  glass 
will  then  be  wet  as  high  as  a  (Fig.  3)  and  the  surface  abc 
will  decrease  in  size  by  assuming  the  shape  a/3c.  In 
doing  this  a  certain  weight  of  liquid  will  be  raised.  When 
the  weight  P  of  this  portion  raised  is  equal  to  the  surface- 
tension  for  that  length  of  surface  equilibrium  will  be 
established,  i.e., 


THE  LIQUID  STATE.  79 

or 


If  now  instead  of  using  a  plate  we  use  a  capillary  tube, 
with  a  radius  equal  to  r,  I  will  become  2nr  and  the  weight 
of  liquid  raised  in  the  tube  will  be 

P  =  2nrx. 
On  the  other  hand  we  have 


where  h  is  the  height  of  the  liquid  in  the  tube  (meas- 
ured downward  in  case  the  liquid  does  not  wet  the  walls), 
and  5  is  the  specific  gravity  of  the  liquid.  We  have  then 


or 


i.e.,  the  surface-tension  of  a  liquid  can  be  obtained  from 
its  specific  gravity  and  the  height  it  ascends  in  a  capil- 
lary tube  of  the  radius  r. 

By  aid  of  the  surface-tension  it  has  been  found  possible 
to  determine  molecular  weights  of  liquids.  Eotvos  found 
experimentally  that 


8o  ZLEMENTS  OF  PHYSICAL  CHEMISTRY. 

where  x  is  the  surface-tension  in  dynes,  and  V  the  rrfolec- 
ular  volume.  Ramsay  and  Shields  expressed  this  rela- 
tion in  the  form 


where  T  is  the  temperature  reckoned  downwards  frojn 
the  critical  one,  d  is  a  constant  equal  to  6,  and  k  is'  a 
constant  independent  of  the  nature  of  the  liquid.  T  here, 
however,  must  always  be  greater  than  35  for  good  results- 
Ostwald  explains  the  relation  as  follows:  If  the  mcje 
were  in  spherical  form,  the  surface  would  be  proportional 
to  the  volume  raised  to  the  power  §,  since 

F:Fi::r3:ri3,    and    s:si  :  ir2:^2. 

We  see,  then,  that  the  above  law  simply  expresses 
the  fact  that  the  work  necessary  to  form  the  molecular 
surface  depends  only  upon  the  temperature,  as  calculated 
downwards  from  the  critical  one  (or  at  least  a  point  about 
6°  below  this),  at  which  the  surface-tension  and  the 
work  are  both  equal  to  zero.  The  law,  in  fact,  is  ex- 
actly analogous  to  the  gas  law  pV  =  RT,  for  at  the  point 
from  which  T  is  reckoned  the  volume  is  equal  to  zero, 
and  the  volume  and  work  above  that  temperature  depend 
only  on  the  temperature  and  are  independent  of  the  nature 
of  the  gas. 

Using  as  M  the  value  found  for  the  gaseous  state 
Ramsay  and  Shields  found  k  for  many  liquids  to  be 


THB  LIQUID  STATE.  81 

equal  to  2.12  ergs.  These  liquids  included  C6H6,  C6H5C1, 
CC14,  C2H6O,  CS2  and  PC13,  and  many  others.  For 
other  liquids  k  was  found  to  be  very  much  smaller  than 
2.12,  which  was  assumed  to  indicate  that  the  V  in  the 
formula  is  too  small,  i.e.,  that  the  molecular  weight,  and 
consequently  volume,  has  been  increased  by  association. 
Thus  for  H2O  k  =  0.9  —  i .  2 ;  for  organic  acids  k = 0.8  —  1.6; 
for  alcohols  &  =  1.0  —  1.6,  etc.  When  K  is  the  value  of  k 
as  found  by  experiment,  Ramsay  and  Shields  assumed  that 
the  factor  of  association  does  not  vary  with  the  tempera- 

2.12  /2.12\f 

ture  and  may  be  found  from  ~jF~=y*>  or^=(  — =-)  , 

where  y  is  the  factor  of  association. 

Later  work  by  Ramsay  and  Aston  has  resulted  in 
substituting  for  the  above  relation  for  y  another  according 

to  which  y  =  i  '  (i  +  fit) )  ,  when  /*  is  a  constant  depend- 
ing upon  the  nature  of  the  liquid.  //  is  not  as  yet  settled 
which  of  these  relations  is  correct,  if  either,  so  that  appar- 
ently it  is  not  yet  possible  to  accurately  determine  y  from 
the  variation  from  the  normal  value  0}  k. 

An  example  of  the  use  of  these  two  equations  will  per- 
haps make  them  clearer.  The  radius  of  a  capillary  tube 
is  0.01843,  at  46°>  ^  (tne  height  of  CC14  in  the  tube)  = 
11.643  cm-j  5  (tne  density)  =  1.5420,  hence  x  =  \  hrs  Xp8o. 
=  22.9  dynes  per  centimeter.  Since  the  critical  tempera- 
ture =283°  and  k  =  2.i2,  x(Mv)$  =  4g2.6  and  M  =  i^. 


82  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

2.12 

Forwaterat  10°  K  =  o.Sjy  hence  ^  =  3.81  (from -^r- =  y$) , 

0.07 

consequently  the  association  is  such  that  the  average  molec- 
ular weight  of  liquid  water  is  3.81X18.  This  could  be 
caused  by  a  mixture  of  associated  and  simple  molecules. 
The  results  given  below  will  show  the  molecular  weights 
of  some  of  the  more  common  non-associated  substances, 
and  the  factors  of  association  of  some  of  those  which  are 
associated,  as  determined  by  both  formulas.  (Ramsay  and 
Shields,  Zeit,  f.  phys.  Chem.,  12,  431-755,  1893);  Ram- 
say and  Aston,  'Proc.  Roy.  Soc.,  56,  162,  1894.) 

NON-ASSOCIATED  LIQUIDS. 
CSg.     Carbon  disulphide. 

t  #(dynes  per  sq.  cm.)  x(Mvft.  yXM 

i9°-4  33-58  515.4 

46°. i  29.41  461.4  1.07X76 

N2O4.     Nitrogen  peroxide. 
i°.6  29-52  461.9 

19°. 8  26.56  423.5  1.01X92 

SiCl4.     Silicon  tetrachloride. 
i8°.9  16.31  383.9 

45°- 5  13-66         •  329.8  i.  06X169.  < 

PC13.     Phosphorus  trichloride. 
i6°.4  28.71  562.3 

46°. 2  24.91  499.8  1.02X137.2 

CC14.     Carbon  tetrachloride i  .01 X  153 .  ( 

C6H0.     Benzene i  .01 X  78 

(C2H5)2O.     Ethyl  ether 0.99X74 

C,H6C1.     Chlorbenzene 1.03X112^ 

C6H5NH2.     Aniline i  .05X93 

C5H5N.     Pyridin 0.93X79 

Hg.*     Mercury  (liquid) 171 

*  Kistjakowski,  J.  russ.  phys.  chem.  Ges.,   34,  pp.   70-90.      This  value  ii 
found  from  surface-tension  assuming  the  critical  temperature  to  be  about  700°  C 


THE.  LIQUID  STATE. 


ASSOCIATED  LIQUIDS  BY  FORMULA  y  =  \~jr-)  • 
ALCOHOLS. 


Name. 

16-46° 
yXM 

46-78° 
yXM 

78-132° 
yXM 

Methvl           

3.43X  32 

3.24X  32 

2.8oX  32 

Ethvf         

2.74X46 

2.43X46 

I    .07X46 

Propvl 

2    25X60 

2    31X60 

Isopropvl 

2    86X60 

2    72X60 

Butyl                           '.  .  . 

I    04X74 

I     72X74 

I    76X74 

Isobutyl  
Arnvl   .  .       

1.95X74 
1.97X88 

1.86X74 
1.69X88 

1.64X74 

I.S7X88 

Allvl 

I    88X58 

I    86X58 

Glvcol 

2    92X62 

2    48X62 

•>   12X6-* 

Formic         

ACIDS. 

3  61X46 

3    13X46 

Acetic       

3  .  62  X  60 

3    t2X6o 

2    77X6O 

Propionic   

.77X74 

.78X74 

88X74 

Butyric 

58X88 

73X88 

69X88 

Isobutyric                   .      .  . 

4.^X88 

82X88 

77X88 

Valerianic  
Capronic      

.36X102 
-46X  116 

.37X102 
47X  116 

.70X102 
49X116 

WATER. 


t 

* 

*U/tO* 

yXM 

0° 
10° 
20° 

30° 
40° 
50° 
60° 
70° 
80° 
90° 

73-21 
71.94 
70.60 
69.  10 

67.50 

65.98 

64.27 

62.55 
60.84 
58.92 

502  .  9 
494-2 

485-3 
476.0 
466.0 
456.7 
446.4 
436.0 

425-9 
414.3 

3.81X18 
3-68X18 
3-44X18 
3.18X18 
3-13X18 
3.00X18 
2.96X18 
2.83X18 
2.79X18 
2  66X18 

100° 

57-15 

403-7 

2.61X18 

110° 
120° 
130° 
I400 

55-25 
53-30 
51-44 
49.42 

392-3 
380.4 
369.2 
356-8 

2.47X18 
2.47X18 
2.32X18 

84 


ELEMENTS   OF  PHYSICAL   CHEMISTRY. 


FACTOR  OF  ASSOCIATION  BY  FORMULA  >'=     -  —  (i+  /*/)  . 


Methyl  Alcohol. 

Ethyl  Alcohol. 

Water. 

/ 

y 

t 

y 

t 

y 

-89.8 

2.65 

-89.8 

2.03 

o 

.707 

+  20 

2.32 

+  20 

1.65 

20 

.644 

70 

2.17 

40 

•59 

40 

.582 

90 

2  .  II 

60 

•52 

60 

•523 

110 

2.06 

80 

.46 

80 

•  463 

I30 

.00 

IOO 

•39 

IOO 

•405 

*5° 

•94 

120 

•33 

120 

•346 

170 

.89 

140 

.27 

140 

.289 

1  80 

.86 

1  60 

.21 

190 

•83 

1  80 

•15 

200 

.81 

200 

.09 

2IO 

.78 

210 

.06 

220 

•75 

22O 

1.03 

230 

I  .00 

Measurements  have  also  been  made  by  Bottomley  in 
this  way  of  the  factors  of  association  of  fused  salts,  using 


,  2.12  if 

the  equation  y  =  ]  —r- 


His  results  are  as  follows; 


POTASSIUM  NITRATE,  KNO3. 


t 

X 

(Mi,)* 

K 

yXM 

338 

406 

109.8 
105.8 

i57i 
1539 

0.471 

9.55X101 

349 
414 

106.4 
loo.  7 

1521 
1485 

0.584 

7.49X101 

341 
407 

106.4 
104.6 

J55i 
I5I9 

0.485 

9.145X101 

THE  LIQUID  STATE. 


SODIUM  NITRATE,  NaNO3. 


f 

X 

(AfiO* 

K 

yXM 

339                  106.4 
405                  101.8 

1374 
1308 

0.500 

8.73X85 

329 
405 

no.  8 
106.5 

1398 
1369 

0.381 

13-13X85 

344 

III  .0 

1403 

°-453 

10.17X85 

For  liquefied  gases  Baly  and   Donnan  found  slightly 
lower  values  for  k  than  2.12.     Their  values  are: 


Substance. 


15 


J: 


Oxygen Y J-91? 

Nitrogen 2 . 002 

Argon 2 . 020 

Carbon  monoxide i .  996 

Oxygen  and  nitrogen  apparently  cut  the  temperature 
axis  at  points  which  are  practically  coincident  with  their 
critical  temperatures,  while  argon  and  carbon  monoxide 
cut  the  axis  at  temperatures  lower  than  their  critical 
values  (9°  for  argon  and  5°  for  carbon  monoxide.) 

For  liquefied  carbon  dioxide  Verschaffelt  has  found  the 
value  £  =  2.222,  and  for  nitrous  oxide,  £  =  2.198,  in  both 
cases  the  temperature  axis  being  cut  6°  below  the  criti- 
cal value. 

These  results,  together  with  the  one  for  liquid  mercury 
given  above,  show  that  this  method  in  its  approximate 
form  (i.e.,  for  molecular  weight  determinations)  is  appli- 
cable to  all  liquids.  It  has  been  shown  by  Ramsay  to 
hold  also  for  liquid  mixtures. 


86 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


24.  Molecular  weight  in  the  liquid  state.  Critical 
temperature. — It  will  be  seen,  thus,  that  in  reality  we 
have  but  two  methods  for  determining  the  molecular 
weight  in  the  liquid  state,  viz.,  the  relations  of  Young  and 
Thomas,  and  that  of  Eotvos  and  Ramsay  and  Shields. 
And  as  the  former  is  only  applicable  to  the  critical  state, 
the  number  is  reduced  to  one  when  any  range  of  tempera- 
ture is  to  be  employed.  It  is  true  that  we  can  tell  from 

div 
the  ratio  -™  for  different  temperatures,  whether  the  ratio 

of  the  molecular  weights  in  the  liquid  and  gaseous  states 
remain  constant,  but  further  than  that  the  relation  is 
valueless  in  this  respect.  In  fact  even  the  ratio  between 
the  two  molecular  weights  is  not  always  easy  of  interpre- 
tation, for  both  may  vary,  as  will  be  seen  from  the  follow- 
ing results  for  acetic  acid. 

FACTORS  OF  ASSOCIATION. 


t 

Liquid  State. 

Gaseous  State. 

20 

2.13 

.98 

40 

2.06 

•87 

60 

•99 

•79 

80 

.92 

•73 

100 

.86 

.70 

120 

•79 

.68 

140 

.72 

.67 

The  relation  of  Trouton  (p.  69)  also  indicates  whether 
the  substance  is  associated  or  non-associated,  but  its 
results  are  not  very  accurate,  so  that  the  method  by  surface- 


THE  LIQUID  STATE.  87 

tension  is  actually  the  only  method  which  is  accurate  and 
applicable  to  any  very  great  extent. 

We  can  now  define  molecular  weight  in  the  liquid 
state ;  in  other  words,  we  can  obtain  a  relation  for  liquids 
which  may  be  used  in  an  analogous  way  to  the  one  com- 
monly employed  for  gases.  It  is  at  once  apparent  that, 
according  to  our  present  use  of  the  term,  the  molecular 
weight  of  a  liquid  at  any  temperature  is  that  weight  in 
grams  which  at  that  temperature  will  occupy  such  a 
volume  V  (in  cc.)  in  the  formula  xV$  =  k(r  —  d)  that  k  is 
equal  approximately  to  2.12  ergs.  For  this  purpose  it  is 
necessary  to  know  the  surface-tension  at  that  temperature 
and  the  term  T,  i.e.,  the  difference  between  the  critical 
temperature  of  the  liquid  and  the  temperature  of  the 
experiment.  As  it  is  not  an  easy  matter  to  determine 
the  critical  temperature  experimentally  this  definition 
is  not  so  convenient  for  use  as  the  one  given  below, 
although  in  essence  both  express  the  same  facts. 

By  the  determination  of  the  surface-tension  and  specific 
volume  of  a  liquid  at  two  different  temperatures,  how- 
ever, the  difficulty  of  the  above  definition  is  avoided, 
and,  at  the  same  time,  it  is  possible  to  determine  the 
critical  temperature  of  the  liquid  with  considerable 
accuracy.  Applying  the  formula  already  used  for  the 
same  liquid  at  two  temperatures,  which  do  not  lie  too  far 
apart,  we  obtain 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 
and 


which  when  combined  give 


where  A(xV*)  denotes  a  finite  change  in  the  term 
corresponding  to  a  finite  change  of  temperature.  In 
this  form  it  is  unnecessary  to  know  the  critical  -temperature- 
of  the  liquid.  By  a  slight  rearrangement  we  obtain 


--2.12, 


i.e.,  the  temperature  coefficient  oj  the  molecular  surface- 
tension  for  all  liquids  is  2.12  ergs.  According  to  this,  then, 
the  molecular  weight  in  the  liquid  state  is  that  weight  in 
grams  which  at  that  temperature  gives  such  a  volume  that 
a  change  in  temperature  of  i°  involves  an  amount  of 
surface-work  equal  to  2.12  ergs.  This  is  similar  to  a 
definition  which  might  be  used  for  the  gaseous  state,  i.e., 
the  one  expressive  of  the  amount  of  external  work  in- 
volved by  the  change  in  temperature  of  i  mole  of  gas 
i°.  This,  according  to  p.  49,  is  equal  to  R,  and  we  may 
say  the  molecular  weight  in  the  gaseous  state  is  that 
weight  in  grams  occupying  such  a  volume  that  a  tempera- 
ture-change of  i°  will  involve  the  external  work  R,  i.e.,  2 
calorics,  0.0821  liter  atmospheres,  or  8.3iXio7  ergs. 


THE  LIQUID  STATE.  89 

As  already  mentioned  this  J-formula  may  be  used  to 
determine  the  critical  temperature.  To  do  this  we 
substitute  for  V  its  value,  assuming  the  gaseous  molecular 
weight,  and  thus  find  the  specific  value  of  k  for  this 
liquid  (i.e.,  the  value  actually  found  in  place  of  2.12). 
Dividing  the  value  of  x(Mv)*  by  this  new  value  for  k  and 
adding  6°  as  well  as  the  temperature  of  observation  to 
this  quotient  we  obtain  the  critical  temperature.  In 
other  words,  we  find  k  by  the  J-formula  and  substitute 
it  in  the  original  form  x(Mv)*  =  k(T  —  d).  An  example 
will  make  this  quite  clear. 

At  i9°-4,  for  CS2,  #  =  33.58,  at  46°.!,  #  =  29.41  dynes. 
We  have  then  for  x(Mv)*  at  the  two  temperatures 


and 

\* 

X  29.41,  —461.4. 


/    76  \* 
—  I 

Vl.22' 


223 

Where  1.264  and  1.223  are  the  densities  at  the  two  tempera- 

d(x(Mv)*)  .  515.4—461.4    54.0 

tures.      v     ,'     is  then  equal  to  — 

J7  46.1  —  19.4       26.7 

2.022,  i.e.,  the  value  for  k  in  this  specific  case.  The  critical 
temperature  of  CS2  is  then  equal  to  ^  -  +6  +  19.4 

=  28o°.3  C.  There  is  a  slight  uncertainty  here  on  account 
of  the  value  6  for  d,  for  this  is  not  the  same  for  all  liquids; 
the  possible  error  is  small,  however. 


po  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

For  T  less  that  35°  the  formula  does  not  seem  to  hold. 
And  this  is  also  true  for  associated  liquids,  only  non- 
associated  liquids  allowing  this  determination  of  critical 
temperature  to  be  made.  Some  of  the  critical  tempera- 
tures determined  in  this  way  by  Ramsay  and  Shields  (I.e.) 
are  given  below  and  will  serve  to  show  the  value  of  this 
method,  which  has  completely  replaced  the  older  and  less 
accurate  relation  of  Schiff,  based  upon  the  empirical 
formulae  xt  =  XQ—fit  and  at2=a^2—^ft,  where  x  is  the 
surface-tension,  a  is  height  of  liquid  standing  in  a  capil- 
lary tube  of  J  mm.  radius,  and  /?  and  /?'  are  the  respec- 
tive temperature  coefficients. 


CRITICAL  TEMPERATURES. 

Substance.  Calculated.      Observed. 

Carbon  disulphide,  CS2  ............   280  .3  275 

273 
277.7 
271.8 
279.6 
Nitrogen  peroxide,  N2O4  ..........  *    226.4  171.2* 

Silicon  tetrachloride,  SiCl4  ........    213-8  230 

221 
Phosphorus  trichloride,  PC13  .......    290  .  5  285  .  5 

Benzene,  C6H6  ...................   288  288.  5 

Chlorbenzene,  C6H6C1  .............   359-7  360 

Ethylacetic  acid,  CH3COOC2H6.  .  .  .    250.5  251 

Ethyl  ether,  (C2H5)2O  ............    191.8  194  .  5 

Carbon  tetrachloride,  CC14  ........   282.  i  283.2 

Paraldehyde,  C6H12O3  .............   310.5 

*  The  calculated  value  here  does  not  account  for  the  dissociation  into 
The  value  171.2  is  the  critical  temperature  of  a  mixture  of  NjC^  and  NOj. 


CHAPTER  IV. 
THE   SOLID    STATE. 

25.  Remarks. — A   solid    is    a    substance    which    pos- 
sesses a  form  of  its  own,  i.e.,  its  shape  is  not  dependent 
upon  that  of  the  vessel  in  which  it  is. 

Just  as  we  find  that  a  substance  contains  less  energy  in 
the  liquid  state  than  it  does  as  a  gas,  so  we  find  that  a 
solid  contains  less  energy  than  a  liquid,  i.e.,  in  going 
into  the  liquid  state,  it  absorbs  heat-energy. 

26.  Atomic  heat.     Law  of  Dulong  and  Petit. — The 
atomic    weight  of  all   elements  have   approximately  the 
same  capacity   for   heat.     A   few   have   been   found   to 
give   different   values,    but   in   general   for   all   elements 
(except  boron,  silicon,  and  carbon)  we  find  that 

atomic  heat  =  atomic  weight  X  specific  heat  =  6.34. 

This  is    the    law  of  Dulong  and  Petit.  By  the  aid  of 

it  it  is  possible  to  determine  the  atomic  weight  from 

91 


92  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

the   experimentally   observed    specific    heat,    i.e.,    for    i 
gram,  we  have 

6.34 

atomic  weight  = r^ — r 7- 7 —          — \  • 

specific  heat  (i.e.,  tor  i  gram; 

According  to  the  law  of  Joule  (1844)  the  heat  capacity 
of  a  compound  is  equal  to  the  sum  of  the  heat  capacities 
of  its  constituents.  The  specific  heat  of  a  compound, 
then,  will  show  the  approximate  number  of  atomic 
weights  of  which  it  is  composed;  and  this  represents  the 
only  method  we  at  present  possess  for  determining  the 
molecular  weight  in  the  solid  state.  In  other  words,  the 
molecular  weight  of  a  compound  in  the  solid  state  may 
be  defined  as  the  weight  which  has  a  heat  capacity  ap- 
proximately equal  to  the  sum  of  the  atomic  heats,  as 
determined  from  the  ratios  found  by  analysis.  This 
definition  is  not  as  convenient  in  form  as  it  might  be,  but 
then  its  importance  is  not  so  great  as  those  for  the  gaseous 
and  liquid  states.  Whether  the  relations  as  expressed 
above  are  correct  or  not  is  impossible  to  say;  at  any 
rate  this  method  is  the  basis  of  the  formula  weights  in 
the  solid  state  as  we  use  them.  An  example  will  serve 
to  make  the  relation  clearer. 

Analysis  shows  that  lead  chloride  contains  103.5  Parts 
of  lead  to  35.45  parts  of  chlorine.  The  question  then 
arises  as  to  the  value  of  the  atomic  weight  of  lead, 
whether  103.5,  207>  or  3IO-55  m  other  words,  is  the 
formula  weight  of  lead  chloride  PbCl,  PbCl2,  or  PbCl3? 


THE  SOLID  STATE.  93 

Since  the  specific  heat  of  lead  chloride  is  0.0664,  accord- 
ing to  the  above  relations,  we  have: 

Molecular          Molecular  Heat  Sum  of 

Weight.  =M X o.o 664.       Atomic  Heats. 

PbCl i°3-5  +  35-45  9-3  2+6.4=12.8 

PbCl2 207.0+70.9  18.5  3X6.4=19.2 

PbCl3 310.5+106.3  27.7  4X6.4  =  25.6 

Since  the  formula  PbC^  gives  a  molecular  heat  ap- 
proximately equal  to  three  times  the  atomic  heat  (6.4), 
this  is  assumed  to  be  the  correct  formula,  i.e.,  the  molec- 
ular weight  is  207.  +  70.9  =  277.9. 

27.  Changes  in  the  state  of  aggregation. — The  most 
important  process  concerning  solids,  for  our  purpose,  is 
their  formation  from  the  liquid  state,  or  the  formation 
of  the  liquid  state  from  them. 

If  a  liquid  is  cooled  very  carefully  it  is  possible  to 
reduce  its  temperature  below  the  solidifying-point  and 
yet  have  no  solid  formed.  If  the  crystal  of  the  sub- 
stance is  thrown  in,  crystallization  takes  place  imme- 
diately, and  the  temperature  rises  rapidly  to  the  true 
solidifying-point.  Besides  this  change  in  temperature  we 
have  also  one  in  volume,  since  the  specific  volume  in  the 
one  state  differs  from  that  in  the  other. 

When  a  gas  condenses  and  goes  into  the  liquid  state 
heat  is  given  off.  In  the  same  way  heat  is  also  given 
off  when  a  solid  is  formed  from  a  liquid,  and  this  is  the 
cause  of  the  rise  in  the  temperature  spoken  of  above. 
The  heat  which  is  liberated  by  a  liquid  solidifying  (or 


94  ELEMENTS    OF  PHYSICAL   CHEMISTRY. 

absorbed  by  a  solid  liquefying)  is  called,  for  i  gram 
of  substance,  the  latent  heat  of  solidification  (or  fusion). 
This  heat,  referring  to  i  mole  of  substance  (heat  for 
the  molecular  weight),  is  called  the  molecular  heat  of 
solidification  (fusion).  The  effect  of  pressure  upon  the 
solidifying-point  may  be  derived  by  the  consideration  of 
equation  (6),  just  as  we  did  for  liquids,  p.  71,  or  in  the 
following  way: 

Starting  with  i  gram  of  a  liquid  at  the  temperature  T, 
occupying  the  volume  v}  and  under  its  own  vapor-pres- 
sure p,  cool  4T°,  i.e.,  to  T  —  4T  and  assume  it  to  be 
completely  transformed  into  the  solid  state.  During  this 
process  an  amount  of  heat  equal  to  w  calories  at  the 
temperature  T  —  AT  will  be  evolved,  the  volume  will 
change  from  v\  to  v2,  the  vapor-pressure  from  p  to  p  —  dp, 
and  an  amount  of  work  equal  to  (v\—v^{p—Ap)  will 
be  involved. 

Now  heat  the  solid  thus  obtained  to  T°  and  allow  it 
to  melt  at  the  pressure  p  to  form  the  original  volume,  v\. 
This  change  in  volume  will  be  the  same  as  the  previous 
one,  except  that  it  has  the  opposite  sign.  The  heat 
absorbed  at  T°  is  so  nearly  equal  to  w  that  the  difference 
may  be  neglected  when  AT  is  small,  and  the  work  in- 
volved is  p(vi—v2).  The  total  work  of  the  whole  pro- 
cess, i.e.,  after  the  original  state  is  once  more  attained,  is 
Jp(vi-v2),  i.e.,  p(vi-v2)-(p-Ap}(vi-v2)J  and  by  the 
second  law  (page  58). 


THE  SOLID  STATE.  95 

(24) 


or 

Jp  _        w  AT     T(vi-v2) 

=  J  =      w 


Vi  being  the  specific  volume  of  the  liquid,  1/2  that  of 
the  solid  substance,  and  w  the  latent  heat  of  fusion  for 
i  gram. 

This  formula  is  not  only  of  use  in  case  of  a  change 
in  the  state  of  aggregation,  but  also  for  any  reversible 
change,  for  example,  that  taking  place  between  allotropic 
forms  of  the  same  substance.  Thus  at  95°.6  C.  the 
rhombic  form  of  sulphur  is  transformed  into  the  mono- 
clinic  form.  The  reaction  is  perfectly  reversible,  an  in- 
crease of  heat  increasing  the  amount  of  the  monoclinic 
form,  a  decrease  increasing  that  of  the  rhombic.  The 
change  in  volume  being  0.014  cm-  and  the  heat  of  trans- 
formation per  gram  2.52  cals.  we  have, 

AT     TJv     (273  +  95.6)0.000014 

4p~    w   '          2.52X0.041  ~°'°5' 

where  the  term  w  is  transformed  into  liter-atmospheres, 
and  the  result  given  in  degrees  per  atmosphere,  i.e., 
an  increase  of  pressure  of  i  atmosphere  increases  the 
transformation  temperature  by  o.°o5,  a  result  which 
agrees  well  with  the  experimental  value. 


96  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

If  the  specific  volume  of  the  solid  is  greater  than  that 
of  the  liquid,  i.e.,  if  the  solid  floats  upon  the  liquid  (ice  in 
water),  the  freezing-point  will  be  depressed  by  pressure. 
If,  on  the  other  hand,  the  specific  volume  of  the  solid  is 
less  than  that  of  the  liquid,  i.e.,  if  the  solid  sinks,  the 
solidifying-point  will  be  raised  by  increased  pressure.  In 
the  former  case  the  pressure  will  tend  to  keep  the  sub- 
stance in  the  liquid  form,  since  that  has  the  smaller  volume, 
and  a  lower  temperature  will  be  necessary  to  produce  the 
solid.  In  the  latter  case  the  formation  of  the  solid  will  be 
aided  by  the  pressure,  and  consequently  the  temperature 
need  not  be  lowered  to  such  an  extent.  This  is  simply 
another  application  of  the  principle  of  Le  Chatelier 
(page  33). 

The  actual  effect  of  pressure  upon  the  solidifying-point, 
however,  is  not  large.  Thus  an  increase  of  pressure  up 
to  8.1  atmospheres  depresses  the  freezing-point  of  water 
but  0.059°  c- 

Equations  (24)  and  (25)  give  us  also  the  heat  of  subli- 
mation, i.e.,  of  the  direct  transition  from  the  solid  to  the 
gaseous  state,  if  the  volumes  represent  those  of  solid  and 
gas;  or  the  volume  of  gas  alone  may  be  used  by  substituting 
for  v  the  value  found  from  pV  =  RT,  changing  w  to  /,  i.e., 
the  gram  heat  to  the  molecular  heat. 

The  sublimation-pressure  of  a  solid  is  to  be  considered 
just  as  the  vapor-pressure  (i.e.,  evaporation-pressure)  of 
a  liquid.  It  is  noticed  that  some  solids  sublime  while 


THE  SOLID  STATE.  97 

others  melt  and  evaporate.  The  reason  for  this  difference 
is  the  fact  that  when  the  sublimation-pressure  exceeds 
the  pressure  of  the  atmosphere  sublimation  takes  place 
and  thus  keeps  the  solid  from  melting.  When  heated 
in  closed  vessels,  however,  the  pressure  becomes  great 
enough  to  prevent  continuous  sublimation,  so  that  after 
a  small  portion  sublimes  and  raises  the  pressure,  the 
remainder  will  melt.  This  can  be  shown  experimentally 
by  heating  iodine,  which  usually  sublimes  and  does  not 
melt,  under  sulphuric  acid,  the  pressure  of  which  prevents 
sublimation  and  allows  fusion  to  take  place. 

Since  at  the  solidifying-  or  freezing-point  the  solid 
and  liquid  exist  in  contact  in  all  proportions,  the  vapor- 
pressure  of  the  solid  must  be  equal  to  that  of  the  liquid, 
both  at  the  same  temperature.  That  this  must  be  so 
follows  from  the  following  process :  Imagine  in  the  annular- 
shaped  vessel,  Fig.  4,  water  at  6,  ice  at  a,  and  the  vapor 


FIG.  4. 

of  the  two  at  c :    all  being  at  the  temperature  of  o°  C. 
If  the  vapor-pressure  over  the  water  at  b  is  greater  than 


9^  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

that  over  the  ice  at  a,  a  distillation  from  b  to  a  must  take 
place.  By  the  evaporation  at  b  heat  will  be  absorbed, 
and  by  the  condensation  of  this  vapor  to  ice  at  a,  under 
the  diminished  pressure,  heat  will  be  liberated.  A  layer 
of  ice  will  thus  be  formed  at  b,  while  the  heat  liberated 
will  melt  an  amount  of  ice  at  a.  We  would  have,  then, 
an  exchange  of  heat  at  the  same  temperature.  If  this 
were  possible,  perpetual  motion  would  be  possible;  since, 
however,  the  latter  is  impossible,  the  vapor-pressure  of  the 
solid  must  be  the  same  as  that  of  the  liquid,  both  having 
the  same  temperature. 

The  latent  heat  of  fusion  of  a  solid  varies  with  the 
temperature  just  as  does  that  of  evaporation,  and  the 
expression  for  it  may  be  derived  in  the  same  manner. 
We  assume  here  that  all  change  takes  place  at  constant 
volume,  i.e.,  that  no  external  work  is  done,  (i)  One  gram 
of  solid  is  allowed  to  melt  at  its  fusion  temperature  T  by 
which  the  heat  absorbed  is  w^,  and  the  resulting  liquid 
heated  to  T  +  t,  which  absorbs  the  heat  at.  (2)  The  gram 
of  solid  is  first  heated  to  T  +  t,  which  absorbs  the  heat 
cst,  and  then  allowed  to  melt,  absorbing  the  heat  wr+f 
Since  the  loss  in  energy  by  the  two  processes  must  be  the 
same,  we  have 


and 


THE  SOLID  STATE.  99 

By  this  we  sec  that  if  the  specific  heat  of  the  solid  is 
larger  than  that  of  the  liquid  state,  the  heat  of  fusion 
decreases  with  increasing  temperature. 

In  the  case  of    H^O  r/  =  i,  ^=0.5;    hence  jy,  =  o.5, 

i.e.,  the  heat  of  evaporation  changes  by  0.5  cals.  for  a 
change  in  temperature  of  i°.  In  this  case  there  is  a 
decrease  in  heat  with  a  decrease  of  temperature,  i.e., 
temperature  and  heat  increase  together.  If  the  sign  of 

Aw 

77^  were  minus  this  relation  would  be  just  the  contrary, 

and  the  heat  would  decrease  with  an  increase  in  tempera- 
ture. 
The  latent  heat  of  fusion  for  water,  then,  is 


Thus  when  freezing  at  —4°,  caused  by  pressure  of  530 
atmospheres,  ww  =  fjS. 

From  the  relation  JT(ci—c8)=w  we  can  find  the 
change  in  temperature  necessary  to  reduce  the  latent 
heat  of  fusion  to  zero,  i.e.,  the  temperature  at  which  it 
passes  through  the  value  zero  and  changes  its  sign. 

The  variation  in  the  latent  heat  of  sublimation  with  the 
temperature  can  be  calculated  by  the  formula  used  for 
the  variation  of  the  heat  of  vaporization,  the  specific  heat 
of  the  solid  being  substituted  for  that  of  the  liquid  (76.) 

Although  in  the  derivation  on  page  94  we  assumed  that 


loo  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

all  the  liquid  solidifies  at  the  temperature  T—AT,  this 
could  not  happen  instantaneously.  There  is,  in  fact,  a 
definite  relationship  between  the  number  of  degrees  of 
overcooling  and  the  fraction  of  liquid  which  as  a  result 
separates  at  once  to  form  solid.  That  this  must  be  sc 
is  obvious  after  a  moment's  thought.  Since  the  freezing- 
point  of  a  liquid  is  that  point  at  which  solid  and  liquid 
can  exist  together  in  all  proportions,  it  must  be  the  final 
point  attained  after  any  overcooling  which  leaves  a 
portion  of  the  liquid.  According  to  this,  then,  only  as 
much  solid  can  separate  after  a  definite  overcooling  a< 
will  just  bring  the  system  from  its  overcooled  point  tc 
its  freezing-point.  Since  the  latent  heat  represents  the 
heat  which  is  evolved  when  i  gram  of  solid  is  formed,  anc 
the  specific  heat  of  the  system  will  regulate  the  rise  ir 
temperature  resulting,  it  is  obvious  that  the  fraction  o 
solid  which  can  be  formed  by  an  overcooling  of  i°  mus 
be  equal  to  the  relation  of  specific  to  latent  heat.  Fo: 
a  greater  overcooling,  then,  we  have 


where  /  is  the  fraction  of  liquid  separating  as  solid.  Thi 
term  c  here,  although  referring  in  reality  to  the  averag< 
specific  heat  of  the  system  liquid-solid  between  the 
temperatures  /  and  At  may  be  replaced  by  the  specific  heai 
of  the  liquid  between  these  limits  ;  and  iv,  which  really 


THE  SOLID  STATE.  101 

refers  to  the  latent  heat  at  the  temperature  /—  At  may  be 
replaced  by  the  latent  heat  at  /°.  This  possibility  is  due 
to  the  fact  that  the  change  in  the  specific  heat  of  the 
system,  as  well  as  the  variation  in  the  latent  heat  per 

degree,  depends  upon  the  specific  heats  ?D.,  the,  two  states 

Jj  "••  '   •  '          *„  »  ' 

so  that  the  ratio  —  remains  constant  iw  matter  whyat  ,th£ 

amount  of  liquid  transformed  into  the  solid  state.  If, 
for  example,  the  fraction  /  of  liquid  solidifies,  owing  to  the 
overcooling  J/,  we  have  as  specific  heat  of  the  system 
(i—f)ci+}ct,  and  the  heat  necessary  to  raise  it  J/°  is 
A[(I—/)CI  +/CJ-  The  latent  heat  at  /—  J/,  however, 
according  to  p.  98,  is  j[w+M(ci—  cs)]  where  /  is  the 
fraction  of  the  amount  of  liquid  becoming  solid  and  thus 
giving  out  /  times  the  latent  heat.  The  decrease  in 
the  amount  of  heat  necessary  to  heat  the  system  is 


the  decrease  in  the  amount  of  heat  delivered  by  the  solid- 
ification is 


and  inspection  shows  tkat  the  two  are  equal,  hence 
the  specific  heat  in  the  liquid  state  and  the  latent  heat 
at  the  freezing-point  may  be  used  throughout  to  give  the 
fraction  of  liquid  separated.  A  specific  example  may 
perhaps  make  this  clearer.  For  water  we  have  ci  =  i, 


102  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

cs=o.5,  while  w  at  o°  is  80  cals.     When  overcooled  to 
—  10°  C.  how  much  of  the  liquid  is  separated  as  ice? 


i.  e.,  125  grams  of  liquid  from  each  liter  will  solidify.     To 

adie.Qk  this.va^lue  we  shall  calculate  the  specific  heat  of 

the  system  and  see  that  the  amount  of  heat  given  out  at 

—  10°  is  sufficient  to  raise  the  temperature  to  o°.     The 

specific  heat  of  the  system  is  (i  —  J)  i  +  JX  .5,  i.e.,  *g*.    The 

latent  heat  at    -10°  is    (80-10X0.5)  =75    cals.      The 
ratio,  specific  heat  to  latent  heat,  then,  as  before,  is 


75     8o' 

The  heat  of  sublimation  (pp.  96  and  99)  at  the  me'  ting- 
point  must  naturally  be  equal  to  the  sum  of  the  heat  of 
fusion  and  the  heat  of  evaporation,  i.e., 


where  p  is  the  vapor-pressure  of  the  solid. 
But 


where  P  is  the  vapor-pressure  of  the  liquid. 


THE  SOLID  STATE.  103 

By  combination,  then,  we  obtain,  by  aid  of  (9), 


T  AT      $  \AT  AT' 

where  $  is  the  vapor-pressure  for  both  liquid  and  solid 
at  the  melting-point.  As  an  example  of  the  use  of  this 
formula  we  shall  calculate  the  difference  in  the  tem- 
perature coefficients  of  the  vapor-pressure  in  the  solid 
and  liquid  states.  For  benzene  at  the  fusion  temperature 
5°.6  the  heat  of  fusion  is  30.18  cals.  and  p  is  equal  to  35.5 
mm.,  and  we  have 

RT*/A       AP 


Ap     AP    30.18X35-5X78 
~Af~AT  =    2(273  +  5.6)2      = 

while  by  experiment  it  is  found  to  be 

Ap     AP 

~       =  2'428~  I'9°5  =°'523 


CHAPTER  V. 
THE   PHASE   RULE.* 

28.  Object  of  the  phase  rule. — Thus  far  we  have  con- 
sidered the  three  possible  states  of  aggregation,  gases, 
liquids,  and  solids,  as  systems  existing  alone,  and  at  most 
have  studied  the  quantitative  relations  of  these  as  such, 
and  their  mutual  transformations  the  one  into  the  other. 
Although  all  the  quantitative  relations  possible  are  thus 
expressed,  there  are  certain  further  qualitative  relations, 
regulating  the  equilibrium  existing  between  the  states, 
which  have  not  been  tauched  upon.  These  relations  are 
summed  up  in  the  phase  rule,  as  derived  by  Willard  Gibbs, 
which  is  the  subject  of  this  chapter. 

In  order  to  understand  quite  clearly  the  object  of  this 
rule,  the  following  systems  may  be  considered:  When 
water  is  placed  in  the  Torricellian  vacuum  of  a  barometer 
the  mercury  is  depressed  and  the  depression  increases 
with  the  temperature  of  the  water.  At  any  one  tempera- 
ture, however,  the  pressure  is  independent  of  the  amount 
of  water  present.  Water  and  its  vapor,  then,  can  exist 

*  For  further  details  on  this  subject  see  Findlay's  "The  Phase  Rule 
and  its  Applications,"  of  which  I  have  made  free  use  in  this  Chapter. 

104 


THE  PHASE  RULE.  105 

permanently  side  by  side,  i.e.,  in  equilibrium  when,  at  a 
given  temperature,  the  pressure  has  a  definite  value.  If 
a  salt  solution  is  substituted  for  the  water,  however, 
the  relation  is  different.  It  is  true  that  an  increase  of 
temperature  still  increases  the  pressure;  but  the  pressure 
is  no  longer  independent  of  the  volume,  for  the  greater 
the  volume  (i.e.,  the  greater  the  amount  of  water  forming 
water- vapor),  the  smaller  is  the  pressure.  The  presence 
of  solid  salt,  however,  in  contact  with  the  solution,  causes 
the  pressure  to  remain  constant,  irrespective  of  the 
actual  volume. 

Further,  by  cooling  water  ice  is  formed,  and  so  long  as 
this  is  present  with  the  water  the  application  (or  removal) 
of  heat  varies  neither  the  temperature  nor  the  vapor- 
pressure  of  the  system.  Thus  it  is  only  at  one  definite 
pressure  and  one  definite  temperature  that  ice,  water  and 
water- vapor  can  exist  together  in  equilibrium.  In  the 
case  of  a  salt  solution,  on  the  other  hand,  we  may  have 
ice  in  contact  with  the  solution  at  different  temperatures 
and  pressures. 

The  phase  rule  enables  us  to  decide  the  necessary  con- 
ditions for  equilibrium  in  cases  of  this  sort,  as  well  as  in 
all  others  where  an  analogous  equilibrium  is  possible. 

29.  The  phase  rule. — Before  formulating  the  law  itself 
it  will  be  necessary  to  define  the  terms  used  by  it.  A 
phase  is  a  physically  distinct,  homogeneous  and  mechanic- 
ally separable  portion  of  the  system;  it  need  not,  however, 


io6  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

be  chemically  simple.  A  gaseous  mixture  or  a  solution 
may  form  a  phase;  but  a  heterogeneous  mixture  of  solids 
represents  as  many  phases  as  there  are  substances  present. 
The  dissociation  of  calcium  carbonate  thus  gives  two 
solid  phases,  calcium  carbonate  and  calcium  oxide,  and 
one  gaseous  phase,  carbon  dioxide.  It  is  to  be  remembered 
that  but  one  gaseous  phase  may  be  present,  for  all  gases 
are  miscible  in  all  proportions.  The  components  of  a 
system  are  those  constituents  which  in  concentration  can 
undergo  independent  variation  in  the  different  phases. 
Thus,  in  the  above  example,  water,  not  hydrogen  and 
oxygen,  is  the  component  ;  and  although  calcium  carbonate, 
oxide  and  carbon  dioxide  are  constituents  of  the  equilib- 
rium, only  two  need  be  considered  as  components,  for  the 
amount  of  the  other  is  not  independent  of  the  amounts  of 
these,  but  dependent  according  to  the  chemical  reaction 


Thus  taking  CaO  and  CO2  as  the  components,  the  compo- 
sition of  each  phase  can  be  easily  expressed. 

In  general,  the  components  are  to  be  chosen  from  the 
constituents  at  equilibrium,  and  are  chosen  as  the 
smallest  number  of  such  constituents  necessary  to  express 
the  composition  of  each  phase  participating  in  the  equi- 
librium, zero  and  negative  quantities  of  the  components 
being  permissible. 

The  degree  of  freedom  is  the  number  of  variable  factors, 
temperature,  pressure  and  concentration  of  the  com- 


THE  PHASE  RULE.  107 

ponents  which  must  be  arbitrarily  fixed  in  order  that  the 
condition  of  the  system  may  be  perfectly  denned.  Thus 
a  gas  has  two  degrees  of  freedom,  for  it  is  necessary  to 
fix  two  of  the  conditions,  temperature,  pressure  or  volume, 
to  define  it;  a  liquid  and  its  vapor  has  but  one,  for  equi- 
librium at  a  certain  temperature  exists  only  at  a  certain 

* 

pressure;  while  a  system  of  a  liquid,  its  solid  and  gas, 
has  no  degree  of  freedom,  for  equilibrium  can  only  exist 
at  a  certain  temperature  and  pressure,  for  otherwise  one 
of  the  phases  will  disappear. 

Gibbs'  phase  rule  defines  the  condition  of  equilibrium 
by  the  relation  between  the  number  of  coexisting  phases 
and  components.  According  to  it  a  system  made  up  oj  n 
components  in  n  +  2  phases  can  only  exist  when  pressure 
temperature  and  composition  have  definite  fixed  values- 
a  system  oj  n  components  in  n  +  i  phases  can  exist  so  long 
as  only  one  o)  the  factors  varies;  and  a  system  of  n  com- 
ponents in  n  phases  can  exist  while  two  oj  the  factors  vary. 
In  other  words,  the  degree  oj  freedom  is  expressed  by  the 

equation 

P  +  F  =  C  +  2,     or    F=C+2-P 

where  P  designates  the  number  of  phases,  C  the  number 
of  components,  and  F  the  degree  of  freedom. 

Since  in  the  equilibrium  of  CaCOs  we  have  two  com- 
ponents (CaO  and  CO2)  and  three  phases  "(two  solid, 
CaO  and  CaCO3,  and  i  gaseous,  CO2)  the  degree  of 
freedom  is  2^2—3  =  1.  This  is  a  univariant  system, 


108  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

then,  just  as  water  and  water-  vapor  (i.e.,  F=  i  +  2  —  2  =  1) 
and  so  in  each  we  have  a  definite  pressure  for  each  tem- 
perature within  certain  limits.  This  formula  for  the 
degree  of  freedom  is  used  as  a  system  of  classification 
for  equilibria,  and  accordingly  we  have  invariant,  uni- 
variant,  bivariant,  multivariant  systems,  the  members  of 
each  group  behaving  similarly. 

30.  Derivation  of  the  phase  rule.  —  Although  this  deri- 
vation is  not  the  original  one  of  Gibbs,  it  may  serve  to 
show  how  such  a  relation  can  be  deduced  without  hy- 
pothesis and  solely  by  mathematical  reasoning. 

Imagine  an  invariant  equilibrium  consisting  of  y 
phases  of  n  components.  Consider  one  phase  alone. 
This  phase  will  contain,  then,  a  certain  amount  of  all 
the  n  components.  It  may  be  a  gas  or  a  liquid,  for  all 
components  go  into  the  gaseous  form  and  into  solution, 
at  least  to  a  slight  extent. 

In  this  phase  which  we  are  considering  let  the  concen- 
trations of  the  n  components  be  c\c%  .  .  .  cn.  Since 
the  equilibrium  is  invariant,  the  composition  of  this 
phase  will  be  altered  by  a  change  of  concentration, 
temperature,  or  pressure  of  the  system.  A  change  in 
one  of  these  will  cause  a  corresponding  change  in  the 
other  two.  This  we  may  express  by  the  equation 


.  .  .  cn,p,  T=o, 
where  F  is  any  function  of  the  variables. 


THE  PHASE  RULE.  109 

The  composition  of  one  phase,  however,  determines 

that  of  all  others  wm'ch  are  in  equilibrium  with  it;    for 
\ 

all  phases  which  are  in  equilibrium  with  one  must  be 
in  equilibrium  with  each  other,  and  this  is  only  possible 
for  a  certain  ratio  of  concentration  between  the  con- 
stituents. Thus  for  a  certain  liquid  phase  we  have  a 
certain  gaseous  one  in  equilibrium  with  it  and  perhaps 
a  solid.  It  follows,  then,  that  the  composition  of  all 
the  phases  is  a  certain  function  of  the  same  variables. 
For  each  phase,  then,  we  have  an  equation  of  the  form 

F(ClC2    •    •    •    Cn,p,   T)=0. 

Since  there  are  y  phases,  we  must  have  y  equations 
of  this  form.  There  are,  however,  n  +  2  variables  in 
each  equation,  so  that  if  y  =  n  +  2,  i.e.,  if  we  have  two 
more  phases  than  components,  we  may  find  a  certain 
definite  value  for  each  unknown  quantity,  since  we  have 
the  same  number  of  equations  as  of  unknown  quantities. 
In  this  case  there  is  only  one  value  for  each  c\c2  .  .  .  cn, 
p,  and  T  at  which  the  system  may  exist  in  equilibrium. 
When  n  components  are  present  in  n  +  2  phases  we  have 
equilibrium  only  for  a  certain  temperature,  a  certain  pres- 
sure^ and  a  certain  ratio  oj  concentration  oj  the  single 
phases,  i.e.,  n  +  2  phases  of  n  components  can  only  exist 
at  a  certain  point  (transition  point)  in  a  system  of  coordi- 
nates. 

If  one  of  these   values  is  changed,   then  one  phase 


Iio  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

disappears  entirely,  and  we  have  n+i  phases  of  n  com- 
ponents. In  this  case  there  will  be  n  +  2  variables  (un- 
known quantities)  in  n  +  i  equations;  hence  it  is  nol 
possible  to  find  an  absolute  value  for  any  one.  We  find, 
however,  relative  values  of  the  n  +  2  variables  c\c^  .  .  .  cn. 
p,  T,  i.e.,  for  each  value  of  T  we  have  only  a  certain  value 
of  p  and  of  each  of  the  terms  ciC2  .  .  .  cn  at  which 
equilibrium  can  exist.  It  is  necessary  to  have  at  least  n 
components  present  in  order  that  a  system  containing 
n  +  i  phases  may  exist  as  a  univariant  equilibrium. 

Where  we  have  n  +  i  phases  of  n  components,  and  the 
conditions  are  altered,  one  phase  disappears  and  we  have 
n  phases  (equations)  and  n  +  2  variables  (unknown 
quantities),  i.e.,  the  composition  of  the  phases  is  uncertain 
and  we  have  a  divariant  equilibrium.  Such  a  one  is  a 
mixture  of  water  and  alcohol,  i.e.,  two  components  in 
two  phases,  liquid  and  gas.  At  a  low  temperature  we 
shall  have  a  univariant  equilibrium,  for  ice  is  formed 
and  two  components  are  present  in  three  phases. 

31.  The  equilibrium  of  water  in  its  phases. —  Fig.  5 
gives  the  curves  for  water  when  plotted  in  a  system 
of  coordinates  of  which  the  abscissae  are  temperatures 
and  the  ordinates  are  pressures — in  other  words,  the 
pressure  curves  for  water  and  ice,  and  the  p.t  curve 
for  water-vapor.  Along  the  curve  WV  liquid  and  gas 
form  a  univariant  equilibrium.  The  curve  IV  is  made  up 
of  the  values  at  which  gas  and  solid  can  exist  in  uni- 


THE  PHASE  RULE. 


ill 


variant  equilibrium,  i.e.,  it  is  the  vapor-pressure  curve 
for  ice.  The  curve  IW  represents  the  conditions  for 
the  coexistence  of  water  and  ice.  As  the  freezing-point 
is  but  slightly  influenced  by  pressure,  this  curve  forms 


wv 


FIG.  5. 

only  a  slight  angle  to  the  axis  of  pressures  (i  atmos.  = 
depression  of  0.00752°). 

The  point  in  which  the  three  curves  intersect  is  the 
transition- point,  i.e.,  the  triple  point  in  which  all  three 
phases  can  exist  in  equilibrium.  This  is  at  o°  the  pres- 
sure being  4.6  mm.  of  Hg. 

If  we  start  with  the  system  containing  water  and 
vapor  (i.e.,  down  the  curve  WV)  in  a  closed  vessel  from 
which  heat  can  be  absorbed  the  point  of  intersection 
will  be  reached,  part  of  the  water  will  freeze,  and  we  shall 
have  three  phases  of  one  substance,  i.e.,  the  transition- 
point.  If  the  temperature  is  still  decreased  either  the 
liquid  or  the  gaseous  phase  will  disappear.  Which  of 
these  depends  upon  the  ratio  of  volume.  If  the  volume 
of  vapor  is  great  enough,  all  the  liquid  will  freeze  and 


H2  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

the  curve  IV  will  be  followed.  On  the  other  hand,  if 
the  volume  of  liquid  is  large  all  gas  will  disappear  and 
the  curve  IW  will  be  produced,  for  the  amount  of  ice 
which  separates  will  increase  the  pressure  and  thus  cause 
all  the  gas  to  condense. 

In  a  system  of  coordinates,  then,  a  divariant  equilib- 
rium is  represented  by  a  surface,  i.e.,  W,  V,  and  I;  a 
univariant  equilibrium  by  a  line,  as  WV,  IW,  and  IV; 
and  an  invariant  equilibrium  by  a  point,  as  O. 

32.  The  phase  rule  and  the  mdentification  of  basic 
salts. — An  exceedingly  clever  and  most  important  appli- 
cation of  the  phase  rule  to  inorganic  chemistry  has  been 
made  by  Miller  and  Kenrick.*  In  the  identification  of 
basic  salts  we  have  a  problem  which  up  to  the  present 
has  seemed  utterly  impossible  of  accurate  solution.  By 
the  application  by  Miller  and  Kenrick  of  the  phase  rule, 
however,  a  great  field  is  opened  up  and  there  seems  to 
be  no  reason  now  why  this  subject  should  not  be  re- 
moved entirely  from  the  position  of  mystery  which  it 
has  occupied  in  the  past.  Naturally  analysis  alone  can- 
not give  much  aid  in  an  investigation,  for  by  it  it  is  not 
possible  to  tell  whether  a  chemical  individual  is  present 
or  a  mixture  in  varying  amounts  of  two  or  more  chemi- 
cal individuals. 
Applying  the  phase  rule  to  the  formation  of  basic  salts 

*  See  Jour.  Phys.  Chem.,  7,  259,  1903,  of  which  I  have  made  very 
free  use. 


THE  PHASE  RULE.  113 

by  the  action  of  water  on  chloride  of  antimony  or  on  the 
nitrate  of  bismuth,  the  temperature  being  low  and  con- 
stant and  the  pressure  being  atmospheric,  we  know:  (i) 
that  if  the  system  which  consists  of  three  components  has 
arrived  at  equilibrium  not  more  than  three  phases  can 
coexist.  Of  these  the  solution  forms  one,  and  the  pre- 
cipitate must  be  either  one  single  homogeneous  substance 
(one  phase),  or  a  mixture  of  two  phases,  for  instance,  of 
two  basic  salts,  or  of  one  basic  salt  with  the  oxide.  (2) 
That  if  the  observed  difference  in  composition  between 
two  precipitates  by  the  action  of  different  quantities  of 
water  on  the  same  salt  is  due  to  their  being  mixtures  of 
the  same  pair  of  basic  salts  in  different  proportions,  the 
composition  of  the  mother  liquors  will  be  the  same  in  the 
two  cases.  (For  the  solution  must  be  in  equilibrium  with 
both  phases,  whatever  their  actual  amount.)  The  possible 
cases  are  thus  divided  into  three  groups: 

1.  The  solutions  are  identical  in  composition  in  different 
experiments,  while    the   composition   of   the    precipitate 
varies. — The  precipitate  is  a  mixture  of  two  phases. 

2.  The  solutions  differ  in  composition,   but  the  pre- 
cipitates have  the  same  composition. — The  precipitate  is 
a  single  chemical  compound. 

3.  Both    solutions    and    precipitates    vary. — The   pre- 
cipitate  is   a   single   phase   of  variable   composition,   a 
"  solid  solution." 

If  it  were  possible  to  represent  the  compositions  of  the 


H4  ELEMENTS   OF  PHYSICAL  CHEMISTRY. 

solution  by  abscissae,  and  those  of  the  precipitates  b} 
ordinates,  the  results  of  a  series  of  experiments  could  b( 
represented  by  a  curve;  perpendicular  lines  would  ther 
correspond  to  case  i,  horizontal  lines  to  case  2,  anc 
slanting  lines  to  case  3.  In  a  three-component  systerr 
this  is  not  possible  in  general.  In  many  cases,  however 
a  pair  or  pairs  of  components  may  be  found  whose  ratic 
in  the  solution  or  precipitate  changes  whenever  the 
composition  of  the  solution  or  precipitate  changes,  and 
only  then;  and  since  for  the  interpretation  of  the  results 
it  is  only  necessary  to  know  whether  the  composition 
of  solutions  and  precipitates  remains  constant  or  changes 
from  experiment  to  experiment,  it  is  sufficient  to  plot 
these  ratios  instead  of  the  compositions  themselves. 

Since  in  the  case  of  the  action  of  water  on  nitrate  oi 
bismuth  the  precipitates  are  all  of  the  general  formula 
wBi2Os,  N2O5,  nH.2O,  the  compositions  of  the  solu- 
tions, also,  may  be  expressed  in  terms  of  Bi2O3,  N2O5, 
and  H2O,  which  are  the  components  of  the  system.  In 
the  curve  below  the  abscissae  give  the  ratios  between 
N2C>5  and  H2O  in  the  solutions,  the  ordinates  those 
between  Bi2O3  and  N2Os  in  the  precipitates.*  The 
only  basic  nitrates  found  are  the  two  noticed  in  the 
curve  and  Bi2O3-N2O5-2H2O,  and  all  the  others  which 
have  been  supposed  to  exist  were  found  to  be  mixtures  of 
these  in  differing  proportions.  The  vertical  line  corre- 

*  See  Allen,  Am.  Chem.  Jour.,  25,  307,  1901. 


fHE  PHASE  RULS. 


spends  to  mixtures  of  precipitates,  and  the  horizontal  ones 
to  pure  chemical  individuals. 

It  will   be  seen  from  this  brief  sketch  that  the  phase 
rule  itself  is  not  at  all  difficult  to  understand,  but  that 


Bi,0, 

^ToT 

2Bi203.y205+H20 


05-HH,  0 


H20 


005 


010    015 

FIG.  6 


025    030 


its  applications  may  give  trouble,  owing  to  the  difficulty  of 
choosing  the  components  and  other  details.  For  further 
applications  to  metallurgy  and  to  chemistry  the  reader 
must  be  referred  elsewhere,  for  here  it  is  only  a  question 
of  the  general  principles  involved  in  the  law  itself. 


CHAPTER  VI. 
SOLUTIONS. 

33.  Definition  of  a  solution. — A  solution  is  a  homo- 
geneous   mixture    which    cannot    be    separated    into    its 
constituents  by  mechanical  means.     The  power  to  form 
solutions   varies   with   the   state   of   aggregation.     Thus 
for  gases  it  is  unconditioned,  while  for  solids,  the  opposite 
extreme,  it  is  small,  though  still  present. 

34.  Gases   in   liquids. — A    true    solution   here   is   one 
in  which  there  is  no  chemical  reaction  between  the  liquid 
and  the  gas,  i.e.,  the  gas  may  be  expelled  by  heat.     Henry's 
law,  which  was  verified  by  Bunsen,  enables  us  to  find 
the  amount  of  gas  which  will  be  absorbed.     When  a  gas 
is  absorbed  in  a  liquid  the  weight  dissolved  is  proportional 
to  the  pressure  of  the  gas.     Since,  however,  pressure  and 
volume    (at    constant    temperature)    are    inversely    pro- 
portional, the  law  may  also  be  expressed  as  follows:    A 
given  amount  oj  liquid  absorbs  at  any  pressure  the  same 
volume  oj  gas.     If  now  we  introduce  the  idea  of  concen- 
tration  (i.e.,   the  amount  of  substance  divided   by   the 
total  volume),  instead  of  the  pressure,   we  obtain  still 

116 


SOLUTIONS.  II? 

another  form  for  the  law:  The  amount  of  gas  dissolved 
by  a  liquid  is  such  that  there  is  always  a  constant  ratio 
between  the  concentration  oj  the  gas  in  the  liquid  and  in 
the  space  above  it.  This  ratio  does  not  depend  upon  the 
actual  concentrations,  being  the  same  for  all  concentra- 
tions within  wide  limits,  and  its  size  is  dependent  only 
upon  the  nature  of  the  liquid  and  its  temperature. 

If,  instead  of  a  single  gas,  a  mixture  of  gases  is  dis- 
solved, this  law  also  holds,  in  that  each  constituent  is 
absorbed  from  the  mixture  to  the  same  extent  as  if  it  were 
present  alone  at  a  pressure  equal  to  its  partial  pressure 
in  the  mixture. 

Among  the  characteristics  of  a  gas  absorbed  in  a  liquid 
are  the  following:  There  is  less  absorption  in  a  solution 
than  in  a  pure  solvent,  and  addition  of  a  substance  to  a 
liquid  already  containing  gas  forces  this  out.  A  liquid 
is  easily  supersaturated  with  gas,  i.e.,  reduction  of  the  pres- 
sure does  not  cause  the  gas  to  evolve  immediately.  The 
presence  of  dust  or  another  gas,  porous  solids  enclosing 
air  and  free  from  the  gas  in  question,  all  cause  the  gas 
to  be  evolved. 

When  a  gas  is  absorbed  in  a  liquid  there  is  always  a 
change  of  volume  in  the  latter.  As  a  general  rule  the 
less  compressible  a  gas  is,  the  greater  is  the  increase  of 
volume  caused  by  it.  The  increase  of  volume  caused 
by  the  solution  of  a  gas  has  been  found  to  be  approx- 
imately equal  to  the  value  b  in  the  equations  of  Budde 


ii 8  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

and  Van  der  Waals  (pp.  34,  35),  as  is  shown  for  severa] 
gases  in  Table  II. 

TABLE  II. 

Gas.                             Av  b 

O                         0.00115  0.000890 

N                         0.00145  0.001359 

H                      0.00106  0.000887 

35.  Liquids  in  liquids. — We  have  in  this  case  three 
possibilities : 

1.  No  solubility,  i.e.,  the  formation  of  two  layers  and 
no  homogeneous  mixture. 

2.  A  mutual  solubility,  but  not  in  all  proportions  (as 
water  in  ether  =  3%  and  ether  in  water  =  10%). 

3.  The  same  relation  as  with  gases,  i.e.,  a  solubility  in 
all  proportions  (as  water  and  ethyl  alcohol). 

The  volume  of  liquid  mixtures  is  not  an  additive  pro- 
perty. The  volume  is  never  equal  to  the  sum  of  those 
of  the  constituents.  Almost  all  mixtures  possess  a  volume 
smaller,  but  a  few  larger,  than  the  sum  of  the  partial 
volumes.  This  difference  is  probably  due  in  some  to  a 
mutual  change  in  the  molecular  weights,  and  in  others 
to  the  change  in  the  molecular  weight  of  one  of  the 
liquids  by  the  other. 

The  most  important  point  for  us  to  consider  here  con- 
cerning these  mixtures  is  the  manner  in  which  they  distil. 
We  shall  consider  first  (i)  cases  where  two  completely 
immiscible  liquids  are  present;  next  (2)  those  which 
have  a  mutual  solubility;  and  finally  (3)  those  for  which 
the  mixture  is  homogeneous. 


SOLUTIONS.  119 

i.  A  liquid  which  consists  of  two  immiscible  constitu- 
ents, present  in  two  layers,  has  a  vapor-pressure  equal  to 
the  sum  of  those  of  the  constituents.  The  boiling-point 
of  such  a  system,  then,  is  that  temperature  at  which  the 
vapor-pressure  becomes  equal  to  the  pressure  of  the 
atmosphere;  it  lies  still  lower,  therefore,  than  the  boiling- 
point  of  the  lower  boiling  constituent.  This  is  difficult 
to  observe  experimentally  by  direct  heating,  owing  to 
the  bumping.  If,  however,  the  vapor  of  another  boiling 
liquid  is  passed  through  the  mixture  the  whole  is  heated 
evenly  and  to  the  same  degree,  and  the  mixture  distils 
below  the  temperature  at  which  either  constituent  alone 
will  do  so. 

Assume  the  vapor-pressures  of  the  constituents  to  be 
pi  and  p2',  the  volume  of  each  in  the  vapor  which  distils 
over  will  be  proportional  to  these  pressures,  and  the  total 
vapor-pressure  will  be  equal  to  the  sum  of  those  of  the 
constituents  in  the  pure  state.  The  weight  of  the  vapor 
of  each  is  equal  to  its  density  multiplied  by  its  volume, 
or,  by  what  is  proportional  to  it,  its  vapor-pressure.  We 
have,  then,  the  weights  (#1  and  q2)  of  the  constituents  in 
the  vapor  from  the  proportion 

qi:q2:  :pidi'.p2d2. 

If  the  vapor-pressures  of  the  two  constituents  and 
their  weights  hi  the -vapor  are  known  it  is  possible  to 
determine  the  vapor-density  of  the  one  in  terms  of  the 


T20  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

other.     Naumann  was   the   first   to   do   this,   using  the 

equation 

,      ,      ?i    ?2 
dii  d2:  :— '  —  . 

Pi    p2 

2.  In  the  case  of  a  mixture  of  partially  miscible  liquids, 
i.e.,  where  each  is  mutually  soluble  in  the  other,   we 
have  also  a  constant  boiling-point  and  a  constant  dis- 
tillate so  long  as  two  layers  remain.     The  boiling-point 
here  may  be  higher,  equal  to,  or,  as  for  class  i,  lowei 
than  that  of  the  lower  boiling  liquid.     It  cannot,  how- 
ever, be  higher  than  that  of  the  higher  boiling  constituent. 
With  regard  to  the  vapor-pressures  of  the  two  layers 
formed  when  the  two  constituents  are  to  a  slight  degree 
mutually  soluble,  Konowalow  has  shown  that  the  liquid 
B  when  saturated  with  A   has  the  same  vapor- pressure 
as  the  liquid  A  when  saturated  with  B,  both  being  at  the 
same  temperature.     Since  the   same   vapor  is   given  off 
by  both  layers,  the  only  effect  of  the  distillation  is  tc 
change  the  absolute   quantities  of  these,  for  the  portion 
lost  by  one  layer  will  be  absorbed  from  the  other.     When 
but  one  layer  remains  the  liquid  is  identical  with  those 
in  class  3. 

3.  When   the   liquids   mix   in   all   proportions,    or   in 
general  when  there   is   only  one  layer  of  liquid,   then 
it  is  possible  to  make  a  complete  separation  of  the  con- 
stituents by  a  fractional  distillation,  provided  the  vapor- 
pressures  of  the  two  differ.     The  vapor  given  off  by  a 


SOLUTIONS. 


121 


homogeneous  mixture  has  a  different  composition  from 
that  of  the  liquid,  and  its  vapor  pressure  is  no  longer 
equal  to  the  sum  of  those  of  the  constituents;  it  depends 
upon  the  action  of  the  liquids  upon  one  another  and 
upon  the  vapors  given  off. 

If  we  use  a  system  of  coordinates  in  which  the  vapor- 
pressures  are  laid  out  upon  the  axis  of  ordinates,  and 
the  percentage  composition  of  the  liquid  upon  the  axis 
of  abscissae,  we  find  that  the  vapor-pressure  never  reaches 
the  sum  of  those  of  the  constituents,  but  at  times  goes 
even  'below  the  smaller  of  these.  Fig.  7  gives  three 


FIG.  7. 


types  of  curves  thus  obtained,  which  will  illustrate  the 
principle  involved.  We  find  in  I  a  maximum  of  the 
vapor-pressure  corresponding  to  a  certain  strength  of 
solution.  In  II  this  maximum  has  disappeared,  while 
in  III  the  opposite  extreme  is  observed,  i.e.,  a  minimum 
of  the  vapor-pressure  which  corresponds  to  a  certain 
composition. 

I.  Mixtures  which  correspond  to  this  type  possess 
a  maximum  of  the  vapor-pressure  for  a  certain  com- 
position, i.e.,  the  lowest  boiling  liquid  is  a  solution  of 


122  ELEMENTS  OF  PHYSICAL    CHEMISTRY. 

a  certain  strength.  An  example  of  this  is  a  75%  solu- 
tion of  propyl  alcohol  in  water.  This  75%  solution 
will  distil  at  the  lower  temperature  from  any  propyl- 
alcohol  solution  until  one  of  the  constituents,  i.e.,  either 
the  water  or  the  propyl  alcohol,  has  disappeared,  when 
the  other  will  remain  behind  in  the  almost  pure  state. 
Thus  if  we  start  with  a  50%  solution  of  propyl  alcohol 
a  75%  solution  will  be  distilled  until  all  the  alcohol  is 
removed  and  water  is  left  in  the  flask.  If  we  start  with 
a  90%  solution  of  the  alcohol  the  75%  solution  will  be 
given  off  until  all  the  water  has  been  used  up  and  pure 
propyl  alcohol  remains  behind. 

II.  For  mixtures  of  this  type  we  can  make  a  com- 
plete  separation   by   distillation.       The   vapor-pressure 
of  all  mixtures  lies  between  those  of  the  two  constitu- 
ents;   consequently  the  one  with  the  higher  vapor-pres- 
sure will  be  given  off  in  the  almost  pure  state,  leaving 
the  other  behind.     Solutions  of  ethyl  and  methyl  alcohol 
in  water  belong  to  this  class.     The  method  of  calculat- 
ing the  value  of  the  vapor-pressure  in  such  a  case  will 
be  given  later.  . 

III.  Here  we  have  a  minimum  of  the  vapor-pressure 
corresponding  to  a  certain  composition.     A  solution  of 
this   composition  will,   then,   always  be  the  last   to   be 
distilled,  since  its  boiling-point  is  the  highest  of  all  pos- 
sible  mixtures.     This   is   just   the   opposite   of   case   I, 
where  the  solution  of  a  certain  composition  is  given  off 


123 


first.     All   solutions,   for  example,    of   formic   acid   and 

water  will  distil  off  an  amount  of  one  constituent  until 

_ 

the  solution  remaining  contains  70%  of  formic  acid. 
A  90%  solution  will  give  off  almost  pure  formic  acid 
until  just  enough  remains  to  form  a  70%  solution.  A 
50%  solution  will  give  off  water  until  the  70%  solution 
is  left  behind. 

An  interesting  technical  application  of  these  principles 
and  facts  has  been  made  by  Young  (Trans.  Chem.  Soc., 
81,  707,  1902)  in  his  method  of  preparing  absolute  alcohol 
from  strong  spirit.  As  is  well  known  ethyl  alcohol  and 
water  form  a  constant  boiling  liquid  (B.P.  =78°.i5)  of 
a  composition  of  95.57%  of  alcohol  and  4.43%  of  water, 
which  by  distillation  alone  cannot  be  further  separated. 
Instead  of  employing  a  dehydrating  agent  and  distilling 
the  alcohol  alone,  as  has  been  the  common  practice,  he 
mixes  with  the  solution  benzene,  w-hexane,  or  a  similar 
liquid.  The  behavior  of  this  system  on  distilling  can  be 
seen  from  the  table  below: 


Liquid  of  Cconstant  B.P. 

B.P. 

Percentage  Composition. 

Alcohol. 

Water. 

Benzene. 

i.  Alcohol,  water  and  benzene.  .  . 

64°.  85 
68°.  25 
6o°.25 

78°-  15 
78°..  3 

80°.  2 
100°.  0 

18.5 
-32.41 

95-57 

TOO.O 

7-4 

8.83 
4-43 

74-1 
67.59 
91.17 

3.  Water  and  benzene  
4.  Alcohol  and  water  
5.  Alcohol  

6    Benzene 

100 

7    Water 

100 

124  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

The  lowest  boiling-liquid  is  the  ternary  system  W.A.B., 
and,  unless  one  liquid  is  present  in  a  relatively  very 
small  quantity,  it  will  distil  first.  If  there  is  more  than 
sufficient  benzene  to  carry  over  the  whole  of  the  water, 
and  if  alcohol  is  present  in  excess,  the  ternary  mixture  will 
be  followed  by  the  binary  mixture  (A.B.),  and  the  last 
substance  to  come  over  will  be  alcohol.  Owing  to  the 
fact  that  there  is  but  3°.4  difference  between  the  boiling- 
points  of  the  ternary  mixture  (A.B.W.)  and  the  binary 
one  (A.B.),  however,  it  is  impossible  to  get  rid  of  all  the 
water  in  one  distillation.  By  redistilling  the  partially 
dehydrated  alcohol  once  or  twice  with  a  further  quantity 
of  benzene  complete  separation  is  effected.  As  the 
alcohol  itself  need  not  be  distilled,  but  can  be  poured 
directly  from  the  still,  and  as  the  benzene  can  be  recovered 
without  difficulty,  this  method  has  many  advantages. 

36.  Solids  in  liquids. — When  a  soluble  solid  comes 
in  contact  with  a  liquid  it  dissolves.  When  no  more 
substance  is  taken  up  by  the  liquid  at  any  one  tempera- 
ture the  solution  is  said  to  be  saturated  at  that  temperature. 
There  are  two  general  methods  of  making  a  saturated 
solution;  but  in  either  it  is  always  to  be  remembered 
that  in  order  that  a  solution  be  saturated  the  solid  sub- 
stance must  be  in  con'tact  with  it. 

I.  The  substance,  in  granular  form,  is  brought  into 
the  liquid,  and  the  system  agitated  at  a  certain  tempera- 
ture until  no  more  is  dissolved;  or, 


SOLUTIONS.  125 

II.  The  solution  is  made  as  above,  but  at  a  higher 
temperature  than  the  desired  one,  to  which  it  is  reduced 
later. 

Both  these  methods  give  the  same  result  if  properly 
carried  out,  although  the  latter  is  quicker  and  so  to  be 
preferred. 

The  importance  of  the  state  of  the  solid  from  which 
the  solution  is  made  has  recently  been  pointed  out 
by  Hulett  (J.  Am.  Chem.  Soc.,  27,  49,  1905).  I  quote 
from  him:  "The  explanation  of  the  phenomenon" 
(of  the  greater  speed  of  solution  and  greater  solubility 
of  very  small  particles  of  solid)  "is  based  on  the  following 
considerations:  The  boundary  between  a  solid  and  a 
liquid  is  the  seat  of  a  certain  amount  of  energy  due  to  the 
surface-energy  of  the  liquid;  if  this  surface  is  increased 
by  powdering  the  solid,  the  total  surf  ace- energy  is  cor- 
respondingly increased.  Further,  it  is  a  generally  ob- 
served fact  that  the  form  of  a  substance  which  has  the 
greater  free  energy  is  the  more  soluble,  has  the  greater 
vapor-pressure,  and  is  the  least  stable  form,  e.g.,  alkr 
tropic  modifications  of  substances  have  different  solu- 
bilities, and  the  unstable  form  is  always  the  more  soluble. 
This  phenomenon  is  hardly  analogous  to  the  well-known 
behavior  of  liquid  drops  of  different  sizes.  Small  drops 
in  the  vicinity  of  large  ones  grow  small  and  disappear, 
while  the  larger  ones  grow  larger,  and  the  reason  is  quite 
clear.  It  is  known  that  the  curved  surface  of  a  liquid 


126  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

has  a  greater  vapor-pressure  than  a  plane  or  less  curved 
surface,  therefore,  a  distillation  takes  place.  The 
similarity  between  the  vapor-pressure  of  liquids  and  the 
solution-pressure  of  solids  has  suggested  to  some  the 
analogy  between  the  facts  just  mentioned  and  the  be- 
havior of  solid  particles  of  different  sizes  in  contact  with 
the  solution.  But  we  cannot  assume  that  the  surface  oi 
the  particles  of  a  powder  are  curved,  or,  if  that  is  granted, 
we  do  not  know  that  a  curved  surface  of  a  solid  or  a 
sharp  edge  has  a  greater  solution-pressure  than  a  plane 
surface  of  the  same  substance." 

Hulett  has  found  that  a  solution,  of  gypsum  saturated 
at  25°,  containing  2.080  grs.  CaSO4  per  liter  (i.e.,  0.01530 
mole)  will  increase  its  concentration  rapidly  to  a  maximum 
when  shaken  with  powdered  gypsum  and  then  will  decrease 
to  the  original  value  again.  In  one  experiment  the  con- 
tent of  gypsum  reached  2.542  grs.  CaSO4  in  a  liter  in  a 
minute.  The  fine  powder  used  for  this  purpose  was 
found,  after  the  concentration  had  reached  its  original 
point,  to  have  increased  in  the  size  of  its  grain.  The 
smallest  particles  thus  go  into  solution,  produce  the 
supersaturation  and  are  then  deposited  upon  the  others, 
so  that  all  increase  in  size. 

It  has  been  observed  that  crystals  dissolve  more  rapidly 
in  some  directions  than  in  others,  and  it  is  possible  thus  to 
cause  a  circular  disk  of  gypsum  (cut  parallel  to  oio) 
to  become  elliptical  by  the  action  of  dilute  hydrochloric 


SOLUTIONS.  127 

acid.  Hulett  calls  attention,  however,  to  the  fact  that 
solubility  has  nothing  to  do  with  rate  of  solubility,  for 
solubility  is  measured  by  the  concentration  of  the  solution 
which  is  in  final  equilibrium  with  the  solid.  He  found 
that  the  plate  of  gypsum  is  equally  soluble  in  all  directions  > 
and  a  solution  saturated  with  planes  cut  in  one  direction 
is  in  equilibrium  with  those  cut  in  other  directions.  For 
these  reasons  Hulett  advocates  the  preparation  of 
saturated  solutions  from  large  particles,  as  there  is 
then  no  danger  of  supersaturation,  and  the  liquid  need 
not  be  filtered.  In  case  small  particles  are  used  and  a 
supersaturation  feared,  contact  with  the  large  particles 
or  crystals  will  cause  the  equilibrium  to  be  attained 
from  the  supersaturated  side. 

It  is  often  possible  to  cool  a  saturated  solution  (which 
has  been  separated  from  its  solid  phase)  to  such  an  extent 
that  it  is  supersaturated.  Some  solutions  of  this  sort  can 
be  kept  indefinitely,  provided  germs  of  the  solid  phase 
are  excluded.  Such  solutions  are  called  metastable.  On 
the  other  hand,  there  are  some  solutions  in  -which  the 
solid  phase  appears  even  when  its  germs  are  excluded. 
Such  solutions  are  called  labile.  The  distinction  between 
these  two  kinds  of  solution  is  not  so  sharp  as  one  would 
think,  however,  for  by  overcooling  sufficiently  the  meta- 
stable solution  may  pass  over  into  the  labile  state.  In 
general  the  metastable  solution  is  smaller  in  concentration 
than  is  that  which  is  labile.  Increase  of  the  concentration 


128  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

causes  the  metastable  state  to  be  transformed  into  the 
labile;  the  concentration  at  which  this  occurs  is  called 
the  metastable  limit. 

37.  Osmotic  pressure. — If  in  a  tall  jar  we  place  a  layer 
of  pure  water  over  a  colored  solution  we  observe  after 
a  time  that  the  whole  liquid  is  colored.  The  colored 
substance  has  diffused  through  the  liquid  until  a  homo- 
geneous mixture  results.  This  action  shows  that  there 
must  be  a  certain  tendency  or  pressure  which  causes  the 
substance  to  occupy  the  entire  space  available.  This 
pressure  causing  the  substance  to  diffuse  into  the  solvent 
is  called  the  osmotic  pressure  oj  that  solution.  If  a  vessel 
is  provided  with  a  partition  of  such  a  nature  that  it  allows 
free  passage  to  the  solvent,  but  not  to  the  solute,  and  we 
have  water  on  one  side  and  a  solution  on  the  other, 
the  solute  will  exert  a  pressure  upon  it ;  this  is  the  osmotic 
pressure  of  that  solution.  Semipermeable  partitions  of 
this  sort  had  been  found  by  Traube,  and  Pfeffer  perfected 
them  in  such  a  way  that  he  was  able  to  actually  measure 
the  different  osmotic  pressures  by  means  of  a  manometer. 
An  apparatus  by  which  the  principle  of  Pfeffer's  measure- 
ments may  be  understood  is  shown  in  Fig.  8.  The 
cylinder  A  is  made  of  porous  clay,  and  is  designed  to 
support  the  semipermeable  film.  To  produce  this  the 
pores  of  the  cylinder  are  filled  with  a  solution  of  copper 
sulphate.  After  the  excess  of  this  is  removed  the  cup  is 
filled  with  a  3%  solution  of  potassium  ferrocyanide,  and 


SOLUTIONS 


129 


allowed  to  stand  for  a  day  in  a  solution  of  copper  sulphate. 
In  this  way  the  pores  are  filled  with  a  film  of  copper 
ferrocyanide,  which  is  permeable  to  water,  but  not  to 
a  large  number  of  salts.  The  process  of  measuring  is 


FIG.  8. 

as  follows:  The  now  semipermeable  cylinder  is  washed 
out  and  filled  with  the  solution  to  be  measured.  The 
rubber  stopper  CC  is  next  inserted  in  A  in  such  a  way 
that  the  solution  rises  a  short  distance  in  the  tube,  i.e^ 
the  cylinder  is  entirely  filled  with  solution  without  air,  and 
the  tube  connected  to  a  manometer.  The  cork  BB  is  then 
fastened  in  the  vessel  DD,  which  is  filled  with  water. 
The  liquid  rises  slowly  in  the  tube  until  equilibrium  is 
reached,  i.e.,  until  the  resistance  in  the  manometer  is 
.just  equal  to  the  osmotic  pressure.  The  height  of  the 
column  of  mercury  thus  represents  the  osmotic  pressure 
of  the  solution  in  A. 

There  is  a  definite  attraction,   then,  or  an  apparent 
one,  between  the  substance  in  solution  and  the  solvent, 


130  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

and  this  is  of  such  a  nature  that  without  the  partition  it 
would  cause  the  system  to  become  homogeneous.  Whether 
this  apparent  attraction  is  actually  an  attraction  between 
solvent  and  solute,  or  whether  it  is  a  tendency  for  the 
solute  to  become  uniform  in  concentration  in  the  liquid, 
owing  to  forces  existing  in  it,  is  unknown.  As  under 
certain  conditions  a  change  in  the  solvent  makes  no 
difference  in  the  osmotic  pressure  observed  (and  when 
it  does  it  can  be  accounted  for  quite  readily)  it  would 
seem  that  the  latter  supposition  is  the  more  correct. 

However,  whether  this  phenomenon  is  a  pressure  or 
an  attraction  can  have  no  difference  from  our  standpoint, 
for  what  we  shall  designate  as  osmotic  pressure  is  an 
experimentally  observed  fact  which  has  been  given  that 
name.  As  such,  then,  we  can  define  osmotic  pressure 
(as  we  shall  use  the  term)  as  that  pressure  which  will 
prevent  pure  solvent  from  going  through  a  semiperme- 
dble  partition  to  dilute  the  solution  contained  within 
it. 

The  nature  of  the  semipermeable  partition  seems 
to  have  no  effect  upon  the  value  of  the  osmotic  pres- 
sure. That  this  must  be  true  can  be  proven  by  the 
following  process:  Imagine  A  and  B  (Fig.  9)  to  be 
two  unlike  semipermeable  partitions  placed  in  a  cylinder 
of  glass.  Assume  the  cylinder  full  of  solution  and 
lowered  in  a  horizontal  position  in  a  vessel  of  water. 
If  the  pressure  P  is  exerted  by  the  partition  A}  and  p 


SOLUTIONS.  I31 

X^LT'TV-i  *'"• 

(a  smaller  one)  by  B,  then  water  will  go  through  A  until 
the  internal  pressure  P  is  reached.  Since  B  allows 
only  the  pressure  p,  however,  this  pressure  P  can  never 
be  reached,  so  that  we  would  have  a  steady  current 
of  water  going  from  A  to  B  under  the  finite  pressure  P  —  p. 
This,  however,  would  be  a  perpetual  motion,  which  is 


FIG.  9. 

impossible;  hence  the  partition  A  must  give  rise  to  the 
same  osmotic  pressure  as  the  partition  B. 

Osmotic  pressure  is,  then,  independent  of  the  nature 
of  the  solvent  and  the  partition,  the  latter  being  simply 
a  means  of  making  it  visible. 

Pfeffer  measured,  by  means  of  an  apparatus  of  the 
type  of  the  one  described,  the  osmotic  pressures  which 
exist  in  a  large  number  of  solutions  and  found  them 
in  many  cases  to  be  very  great.  A  few  results  for  sugar 
solutions  of  varying  concentrations  at  15°  C.  are  given 
in  Table  III. 


TABLE  III. 


c 

P 

p/c 

c 

P 

p/c 

I 

53-8 

53-8 

4 

208.2 

52.1 

I 

53-2 

53-2 

6 

307-5 

5i-3 

2 

101.6 

50.8 

i 

53-5 

53-5 

2.74 

151.8 

55-4 

132  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

where  c  is  the  percentage  composition  of  the  solution, 
p  is  the  osmotic  pressure,  in  cms.  of  mercury,  and  p/c 
is  the  ratio  of  pressure  to  concentration. 

Considering  the  difficulties  in  measuring  and  the 
imperfections  in  the  semipermeable  film,  the  term  p/c 
is  to  be  considered  as  constant;  hence  the  osmotic  pres- 
sure is  proportional  to  the  concentration  of  the  solute. 
Further  results  at  different  temperatures  showed  that 
the  osmotic  pressure  is  proportional  to  the  absolute  tem- 
perature. This,  however,  is  only  strictly  true  when  the 
heat  of  dilution  of  the  solution  is  zero,  i.e.,  when  the  solu- 
tion is  dilute.  On  account  of  experimental  -difficulties 
the  direct  effect  of  this  heat  of  dilution  upon  the  osmotic 
pressure  of  concentrated  solutions  has  not  as  yet  been 
determined. 

van't  Hoff  in  1887  from  this  much  knowledge  derived, 
by  the  aid  of  thermodynamics,  an  analogy  between  the 
behavior  of  gases  and  substances  in  solution.  Here, 
however,  we  shall  not  go  into  his  reasoning,  but  shall 
consider  how  his  results  may  be  arrived  at  in  a  more 
simple  manner. 

As  we  have  seen  by  Pfeffer's  results,  the  term  p/c 
remains  constant  for  any  one  solution  when  p  is  the 
osmotic  pressure  and  c  the  concentration,  i/c,  however, 
is  equal  to  v,  the  volume;  we  have  then 

p/c=pv. 


SOLUTIONS.  133 

Since  now  the  osmotic  pressure  is  proportional  to  the 
absolute  temperature,  as  is  also  the  term  p/c,  we  have 

pv=  constant  XT. 

This  equation  is  so  much  like  the  one  for  gases  (9) 
that  it  immediately  suggests  that  there  is  some  kind  of  a 
connection  between  them  and  substances  in  solution. 
If  now  the  value  of  the  constant  is  determined,  then 
by  comparing  it  with  the  gas  constant  it  is  possible  to 
find  this  connection.  Since  we  are  to  use  the  molec- 
ular gas  constant  in  the  comparison,  it  will  be  neces- 
sary also  to  find  this  constant  for  i  mole.  A  i%  solu- 
tion of  sugar  at  o°  is  held  in  equilibrium  by  a  column 
of  mercury  49.3  cms.  high  and  thus  gives  an  osmotic 
pressure  equal  to 

49.3X13.59=671  grams  per  sq.  cm. 

Since  the  molecular  weight  of  sugar  (CnH^On)  is 
342,  the  volume  in  which  i  mole  is  dissolved  (to  give  a 
i%  solution)  is  34200  cc.  We  have  then 


PoVo    671X34200 

constant  =  ^=-  =  —  -  =  84200  gr.-cms., 

l  273 

while  the  molecular  gas  constant  is 
#=84800  gr.-cms. 


134  ELEMENTS  OF  PHYSICAL   CHEMISTRY, 

The  constant  is  the  same,  then  (within  the  experi- 
mental error),  for  the  gaseous  state  and  the  state  of  a 
substance  in  solution;  consequently  for  each  state,  con- 
sidering i  mole,  we  have 

pV=RT. 

van't  Hoff  summed  up  these  results  in  the  following 
form:  The  osmotic  pressure  of  a  substance  in  solution  is 
the  same  pressure  which  that  substance  would  exert  were 
it  in  gaseous  form  at  the  same  temperature  and  occupy- 
ing the  same  volume. 

Since  i  mole  of  gas  at  o°  and  76  cms.  of  Hg  occupies 
the  volume  of  22.4  liters,  at  the  volume  of  i  liter  it  will 
have  the  pressure  of  22.4  atmospheres  (Boyle's  law). 
This  pressure,  22.4  atmospheres,  is  then  the  osmotic 
pressure  which  is  exerted  by  i  mole  of  solute  in  i  liter 
of  solution  at  o°,  i.e.,  by  a  molar  solution.  From  this  os- 
motic pressure,  22.4  atmospheres,  it  is  very  easy  to  find 
that  for  any  other  concentration  given  in  terms  normal. 
Thus  a  n/io  solution  gives  an  osmotic  pressure  of  2.24 
atmospheres.  That  within  the  limits  thus  far  possible  the 
osmotic  pressure  is  approximately  equal  to  that  pressure 
which  would  be  exerted  if  the  substance  were  in  the 
gaseous  state  in  the  same  volume,  at  that  temperature, 
is  not  a  hypothesis,  but  an  experimental  fact.  It  is 
unfortunate  that  as  yet  osmotic  pressures  cannot  be 
measured  accurately  above  a  few  atmospheres.  From 


SOLUTIONS.  135 

the  advance  which  has  been  made  lately  by  Morse  and 
Frazer  (Am.  Chem.  J.,  25,  i,  1903),  it  is  to  be  hoped 
that  we  shall  soon  be  able  to  test  this  at  present 
limited  law  between  wide  limits.  From  preliminary 
results  thus  far  obtained,  however,  more  concentrated 
solutions  seem  to  give  too  high  a  value  by  experiment, 
just  as  is  observed  with  the  gas  laws  themselves. 

This  law  of  van't  HofFs  enables  us  at  once  to  define 
the  molecular  weight  of  a  dissolved  substance,  and  the 
definition  naturally  reminds  one  of  the  definition  for 
substances  in  the  gaseous  state.  The  molecular  weight 
of  a  substance  in  solution  is  the  weight  in  grams  which 
in  the  volume  of  approximately  22.4  liters  of  solvent  will 
give  an  osmotic  pressure  of  i  atmosphere  at  o°  C.,  or  a 
corresponding  value  at  another  temperature  or  volume. 

If  the  osmotic  pressure  of  a  solution  is  known  it 
is  then  possible  to  find  from  it  the  molecular  weight. 
A  2%  solution  of  sugar  gives,  at  o°  C.,  an  osmotic  pres- 
sure equal  to  101.6  cms.  of  Hg.  If  this  were  a  molar 
solution  it  would  give  a  pressure  equal  to  22.4X76 
=  1702.4  cms.  of  Hg.  The  2%  solution  contains  20  grams 
in  the  liter;  hence  20^1:101.6:1702.4;  ^  =  335. 

The  determination  of  the  osmotic  pressure  directly 
by  the  use  of  Pfeffer's  apparatus  is  not  very  satisfac- 
tory, owing  to  the  difficulty  in  preparing  the  semiper- 
meable  film  and  to  the  fact  that  this  latter  is  easily  broken 
down  so  that  liquid  goes  through.  Morse  and  Frazer 


I36  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

have  succeeded  in  obtaining  electrolytic  films  which 
have  stood  high  pressures,  and  have  obtained  pressure 
as  high  as  31.4  atmospheres  for  molar  and  13  to  14  atmos- 
pheres for  half -molar  sugar  solutions  at  25°  C.  without  be- 
ing able  to  detect  any  breaking  down  of  the  film.  In  the 
absence  of  any  such  method,  however,  it  is  usual  to  cal- 
culate the  osmotic  pressure  from  the  molar  (i.e.,  moles  per 
liter)  concentration  of  the  solution,  as  observed  by  other 
methods  given  later,  i.e.,  to  assume  that  the  general 
law  would  hold  if  the  osmotic  pressure  could  be  accu- 
rately determined.  It  is  not  only  water  solutions  that 
have  been  measured,  however,  for  Raoult  (Zeit.  f.  phys. 
Chem.  27,  737,  1895)  has  found  that  vulcanite  allows 
passage  to  ether  but  not  to  methyl  alcohol.  Thus 
with  ether  on  one  side  and  an  equal  volume  mixture  of 
ether  and  methyl  alcohol  on  the  other,  he  observed  an 
osmotic  pressure  of  50  atmospheres  at  13°  C.,  when  the 
manometer  was  fractured.  In  this  case  the  pressure,  just 
as  with  water  solutions,  was  exerted  in  the  solution,  i.e., 
the  pure  solvent  goes  toward  the  solution  and  a  pressure 
is  necessary  to  prevent  it. 

There  is  one  characteristic  of  osmotic  phenomena 
which  must  be  mentioned.  It  is  the  slow  rise  of  the  pres- 
sure up  to  its  maximum  value.  Thus  with  a  half-molar 
sugar  solution  Morse  found  95  hours  necessary  for  the 
establishment  of  the  final  equilibrium.  This,  naturally, 
is  to  be  attributed  to  the  resistance  of  the  cell  to  diffusion. 


SOLUTIONS.  137 

A  simple  and  quick  method  of  observing  the  effect  of 
osmotic  pressure  is  given  by  Tammann.  If  in  a  moder- 
ately strong  solution  of  copper  sulphate  we  place  a  drop 
of  a  strong  solution  of  potassium  ferrocyanide  it  is  imme- 
diately surrounded  by  a  semipermeable  film  of  copper 
ferrocyanide.  Since  the  ferrocyanide  solution  is  more 
concentrated  than  that  of  the  copper  sulphate,  water  goes 
through  the  film,  from  the  copper  salt  to  the  ferrocyanide, 
and  dark  streaks  are  observed  to  sink  from  the  bubble. 
These  streaks  are  of  stronger  copper-sulphate  solution, 
which  is  formed  from  the  other  by  the  loss  of  water,  and 
sink  on  account  of  their  increased  specific  gravity.  If 
the  ferrocyanide  solution  is  weak  and  another  salt  or 
more  ferrocyanide  is  added  to  it  until  no  streaks  are 
observed  to  sink  from  a  bubble,  then  there  must  be  an 
equal  number  of  moles  per  liter  in  both  of  the  two  solu- 
tions, i.e.,  in  the  copper  solution  and  the  ferrocyanide 
plus  salt.  If  the  ferrocyanide  solution  is  weaker  than 
the  one  of  copper  sulphate  (in  moles  per  liter),  then  water 
will  go  out  of  the  bubble,  which  will  decrease  in  size 
and  light  streaks  will  appear  in  the  copper  solution. 

In  this  experiment  the  tendency  for  the  solutions  to 
become  of  the  same  molar  concentration  is  particularly 
striking,  and  water  is  always  observed  to  go  from  the 
more  dilute  to  the  more  concentrated,  expressed  in 
moles  per  liter.  For  later  work  see  Appendix. 

Consider  a  cylinder,  as  shown  in  Fig.  10,  filled  with 


138 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


a  solution,  and  provided  with  a  semipermeable  piston 
a.  If  by  a  certain  weight  this  piston  is  lowered,  water 
will  go  through  it,  and  we  shall  have  separated  an  amount 


Solution 


Salt 


a 


FIG.  10. 

of  solvent  from  the  solution.  If  the  amount  of  water 
separated  previously  contained  i  mole  of  solute,  and  the 
total  volume  is  very  large,  the  osmotic  work  is  given  by 
the  equation 

pV=RT, 

where  V  is  the  decrease  in  volume  occupied  by  the  solute 
under  the  constant  osmotic  pressure  p.  It  is  possible 
thus  to  determine  the  molecular  weight  of  any  substance 
in  solution  by  finding  the  work  necessary  for  the  separa- 
tion of  a  certain  amount  of  the  solvent.  For  each 
amount  which  has  contained  one  mole  of  solute  the 
work  is  (just  as  for  gases)  2.T  cals.,  so  that  the  number  of 
moles  in  the  definite  amount  of  solvent  removed  may  be 
calculated  from  the  work,  and  from  this  the  molecular 


SOLUTIONS.  139 

weight  of  the  solute  can  be  found.  The  molecular  weight 
of  a  substance  in  solution  by  this  method  would  be  the 
weight  which  removed  from  a  solution  against  its  osmotic 
pressure  would  require  the  work  of  2T  cals.;  or  what  is 
the  same  thing,  the  weight  which  has  been  dissolved  in 
the  volume  of  solvent  requiring  the  work  2 T  cals.  to  re- 
move reversibly  from  the  main  portion  of  the  liquid. 

38.  Electrolytic  dissociation  or  ionization. — The  mo- 
lecular weights  of  a  very  large  number  of  organic  sub- 
stances in  water  are  found  correctly  from  the  osmotic- 
pressure,  i.e.,  they  correspond  to  the  values  in  the  gaseous 
state.  For  other  substances,  however,  varying  results  are 
obtained.  The  osmotic  pressure  is  here  always  too  large, 
and  consequently  from  our  definition  the  molecular  weight 
is  smaller  than  that  observed  in  the  gaseous  state.  This 
reminds  one  immediately  of  the  abnormal  densities  of 
gases.  In  the  case  of  the  gas  the  volume  increases  and  the 
pressure  is  constant;  in  that  of  the  solution,  however, 
the  volume  remains  constant,  while  the  pressure  changes. 

Arrhenius  in  attempting  to  find  an  explanation  for  the 
abnormal  osmotic  pressures  found  by  experiment  that 
those  substances,  and  only  those,  which  give  abnormal 
osmotic  pressures  in  solution  are  capable  of  conducting 
the  electric  current,  and  if  they  are  dissolved  in  other 
solvents  in  which  they  behave  normally  they  lose  this  power. 

Arrhenius  determined  the  electrical  conductivity  of 
the  substance  in  terms  of  molecular  conductivity.  The 


*4°  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

molecular  conductivity  of  a  solution  may  be  denned  as 
the  reciprocal  of  the  resistance  (in  ohms)  of  the  volume 
of  liquid  which  contains  one  formula  weight  of  the 
substance,  the  electrodes  being  i  cm.  apart  and  large 
enough  to  contain  between  them  the  entire  amount  of 
solution.  This  value,  naturally,  is  not  found  directly, 
but  is  calculated  from  that  value  found  for  a  centimeter 
cube  of  the  solution.  (See  Chapter  IX.)  In  this  way, 
always  having  i  mole  between  the  electrodes,  he  found 
that  the  more  dilute  the  solution  the  greater  is  the  mo- 
lecular conductivity.  In  many  cases,  indeed,  he  was 
able  to  reach  such  a  dilution  that  the  molecular  conduc- 
tivity attained  a  maximum  value,  which  is  unaffected 
by  further  dilution.  This  molecular  conductivity  at 
infinite  dilution,  as  it  is  called,  is  designated  by  the  term 
jw^,  that  value  for  any  dilution  F,  being  designated 

by/V 

From  this  it  is  apparent  that  the  solution  undergoes 
some  kind  of  a  change  as  the  result  of  dilution;  and  "the 
investigation  of  such  solutions  at  various  dilutions  shows, 
indeed,  that  the  molecular  weight  (according  to  definition) 
also  changes  with  the  dilution;  the  molecular  weight 
decreasing  to  a  minimum  constant  value,  which  for  binary 
electrolytes  is  one-half  the  formula  weight  of  the  substance 
dissolved.  It  is  not  unreasonable,  then,  to  infer  that 
the  breaking  down  of  the  molecular  weight  is  the  factor 
which  causes  the  conduction  of  the  electric  current. 


SOLUTIONS.  141 

These  facts  formed  the  starting  point  of  what  is 
known  at  present  as  the  "theory  of  electrolytic  dissoci- 
ation." 

As  this  theory  to-day  is  much  misunderstood  by  many, 
and  is  the  subject  of  much  speculation  on  the  part  of 
others,  it  will  be  necessary  for  us  to  consider  carefully 
just  what  is  fact  and  what  assumption,  and  to  see  clearly 
which  portions  are  hypothetical,  and  which  are  destined 
to  remain  under  any  hypothesis  or  lack  of  hypothesis; 
in  other  words,  which  are  experimental  facts.  It  may  be 
said,  however,  that  that  which  is  hypothesis  in  this  theory 
is  unessential,  as  far  as  the  use  of  the  data  is  concerned, 
and  the  only  hypothesis  present,  as  we  shall  consider  it, 
is  that  inherent  in  the  terminology,  which  is  a  relic  of  the 
atomistic  hypothesis  and  utterly  beyond  our  power  either 
to  prove  or  disprove. 

The  salient  facts  which  have  been  grouped  in  this 
theory,  for  it  is  a  theory  in  the  sense  that  it  is  a  law  of 
nature  holding  between  certain  limits,  although  these  are 
not  as  yet  definitely  fixed,  are  as  follows: 

(1)  The  molecular  conductivity  of  certain  substances 
in  water  is  found  to  increase  up  to  a  maximum,  constant 
value,  and  this  increase  is  the  result  of  dilution. 

(2)  Those   solutions   which   conduct   the   current   also 
give   abnormal   osmotic   pressures;   in   other   words     the 
molecular    weight    (as    defined    above)    decreases    with 
increased  dilution  and  finally  reaches  a  minimum  value, 


142  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

which,   for  binary  electrolytes,   is  one-half  the  formula 
weight  of  the  substance. 

(3)  Those    substances    which    in    water    conduct    the 
current  and  give  abnormal  osmotic  pressures,  give  normal 
osmotic  pressures  when  dissolved  in  other  solvents   in 
which  they  do  not  conduct. 

(4)  The  nearer  the  value  of  pv  is  to  that   of  //«>,  the 
more  abnormal  the  value  of  the  osmotic  pressure  (mo- 
lecular weight)   of  the  solution.    And  the  solution  for 
which   /*«>    is   found   also   gives   the    maximum  osmotic 
pressure,    i.e.,  the  minimum  molecular  weight. 

(5)  The  molecular  conductivity  of  a  solution  at  infinite 
dilution  is  an  additive  value,  i.e.>  is  equal  to  the  sum  of 
the  conductivities  of  the  substances  of  which  it  is  com- 
posed.    The  meaning  of  this  is  as  follows:   The  molecu- 
lar   conductivity    at    infinite    dilution    of,    for    example, 
potassium  chloride  plus  that  of  nitric  acid  minus  that  of 
potassium  nitrate  is  found  to  be  equal  to  that  of  hydro- 
chloric acid.     In  other  words, 


For  this  to  be  true,  and  it  is  true  in  general  for  all 
substances,  it  is  necessary  that  the  molecular  conduc- 
tivity of  each  substance  in  solution  be  the  sum  of  two 
values  which  are  independent  each  of  the  other.  Chlorine, 
for  example,  as  the  constituent  of  an  electrolyte,  at  the 
dilution  giving  ,««,,  has  the  same  conducting  effect  when 


SOLUTIONS.  H3 

part  of  a  compound  with  one  element  as  it  has  when 
combined  with  any  other.  It  is  possible,  then,  to  find 
the  value  of  //«>  for  any  binary  electrolyte  when  the 
values  for  the  elements  composing  it  are  known.  In 
other  words,  the  conductivities  of  the  solution  as  pro- 
duced by  the  presence  of  any  element  can  be  calculated; 
and  from  these  values,  by  summation,  the  value  of  //<» 
for  any  binary  electrolyte  can  be  found. 

6.  When   a  solution   is   electrolyzed,   the  products  of 
electrolysis   appear   instantaneously  at  the   electrodes   so 
soon  as  the  circuit  is  completed.     This  indicates  (since 
the  solvent,  water,  does  not  conduct  beyond  a  very  small 
extent)    that   whatever  does  carry  the  current   through 
the  liquid   is   charged  with  electricity  even   before   the 
current  is  applied,  for  the  conduction  is  due  to  the  dis- 
solved substance.     (See  Chap.  IX.)    Further,  it  is  observed 
that  the  same  amount  of  electricity,  96537  coulombs,  is 
necessary  for  the  separation  of  one  equivalent  weight 
(in  grams)  of  any  element;    in  other  words  that  96537 
coulombs    of    electricity    are    transported    through    the 
liquid   with   each   equivalent   weight    (in   grams)    of  an 
element.     (Faraday's  Law,  see  Chapter  IX.) 

7.  The  properties  of  electrolytes  are  found  to  be  the 
sum  of  the  properties  of  the  products  observed  during 
electrolysis.     Thus  any  solution  giving  off  chlorine  on 
electrolysis,  excluding    secondary  reactions,  will  precipi- 
tate  silver  from   its   solution   as   the   chloride.     And   if 


144  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

chlorine  cannot  be  produced  in  any  way  by  the  elec- 
trolysis, silver  will  not  be  precipitated  as  chloride  from 
its  solutions.  And,  on  the  other  hand,  silver  is  only 
precipitated  by  chlorine  when  contained  in  a  solution 
from  which  silver  can  be  deposited  by  the  current  by 
primary  action. 

The  catalytic  effect  of  acids  on  the  inversion  of  sugar 
as  well  as  on  the  decomposition  of  methyl  acetate,  is 

found  to  be  proportional  to  the   ratio  —  for  the   acid; 

/ioo 

and  when  a  large  amount  of  a  salt  of  this  acid  is  added 
to  the  acid  this  effect  is  decreased.  But  this  is  only 
true  when  the  salt  added  is  an  electrolyte. 

All  copper  solutions,  when  very  dilute,  show  the  same 

blue  color,  and  this  also  depends  upon    the  ratio  — p, 

,  /*00 

and  can  also  be  changed,  as  the  effect  of  acids  were 
above,  by  the  addition  of  a  large  amount  of  an  elec- 
trolyte which  contains  the  same  acid  radical  as  the  cop- 
per salt  in  question. 

8.  Observation  shows  that  when  an  element  is  sepa- 
rated on  one  electrode,  anode  or  cathode,  it  is  always 
separated  on  that  one  by  primary  action ;  in  other  words, 
the  sign  of  the  electricity  transported  by  an  element  is 
always  the  same.  Unless  an  element  in  the  pure  state, 
when  dissolved  in  water,  reacts  with  the  water  it  does 
not  conduct  the  current.  This  circumstance  is  assumed 
to  be  due  to  the  fact  that  only  one  kind  of  electricity 


SOLUTIONS.  145 

could    be  carried  by  the  substance,  and  hence    it  pro- 
duces no  conduction. 


The  question  now  arises  as  to  what  theory  can  be 
found  to  correlate  these  facts  and  observations  so  that 
the  generalization  thus  obtained  may  be  employed  to 
foresee  other  facts,  and  applied  to  other  observations 
that  they,  in  their  turn,  may  be  elucidated  and  general- 
ized. By  the  word  theory,  then,  we  do  not  mean  a 
hypothesis,  in  which  something  not  observed  is  added 
to  the  facts  to  "  explain"  them,  but  only  a  generalization 
of  observed  facts.  In  other  words,  what  law  of  nature, 
holding  within  definite,  if  small,  limits,  can  be  obtained 
from  the  above  experimental  facts  when  considered  to- 
gether? 

The  generalization  which  has  been  made  from  these 
facts  is  known  as  the  theory  of  electrolytic  dissociation, 
and,  considering  those  portions  which  are  free  from 
hypothesis  and  fulfil  the  above  conditions,  in  other  words, 
omitting  the  hypothetical  portions  which  it  has  attained 
since  the  time  of  its  inception,  we  find  in  it,  within  cer- 
tain limits,  a  definite  law  of  nature. 

The  principal  points  of  this  theory  are  summarized 
below  in  brief  form,  and  will  each  be  expanded  in  the 
later  portions  of  the  book. 


146  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

That  a  substance  in  solution,  which  conducts  the 
electric  current,  is  dissociated  or  ionized  into  its  con- 
stituents, and  these  constituents,  when  secondary  actions 
are  excluded,  appear  at  the  electrodes  during  elec- 
trolysis. The  extent  of  the  ionization  or  dissociation 
in  any  solution  being  given  at  the  dilution  V  (number 
of  liters  in  which  i  mole  is  dissolved)  by  the  ratio 


That  the  products  of  ionization  or  dissociation  are 
charged  with  electricity,  96537  coulombs  being  carried 
by  the  gram  equivalent  of  any  element  (see  (6)  above). 
A  further  proof  of  this  charged  state  of  ionized  matter 
is  given  by  the  fact  that  not  only  is  the  current  carried 
by  a  solution  dependent  upon  the  number  of  gram  equiva- 
lents transported,  but,  as  we  shall  see  later,  any  other 
means  of  depositing  the  constituents  of  the  solution 
upon  the  electrodes  liberates  an  amount  of  electricity 
which  depends  also  upon  the  number  of  gram  equiva- 
lents deposited.  And  all  cells  in  order  to  give  a  current 
must  contain  electrolytes,  i.e.,  solutions  which  are  ionized. 

Since  a  solution  which  by  conductivity  is  shown  to  be 
completely  ionized,  or  practically  so,  leads  to  a  molecu- 
lar weight,  by  osmotic  pressure  or  analogous  methods, 
of  one-half  the  value  expressed  by  the  formula  weight, 
then,  from  the  case  of  hydrochloric  acid  in  solution,  where 

I  HC1=H'+C1', 


SOLUTIONS.  H7 

the  molecular  weight  of  the  hydrogen  and  chlorine  in 
the  ionic  state>  according  to  our  definition  of  molecular 
weight,  must  be  synonymous  with  the  atomic  weight. 

The  ionic  state,  then,  is  an  allotropic  form  of  the 
ordinary  state  of  the  constituents;  it  differs  from  that 
in  being  charged  with  electricity,  having  less  energy 
than  when  in  the  gaseous  state,  and  always  being  trans- 
formed into  the  ordinary  state  on  the  loss  of  its  charge 
of  electricity. 

Since  the  constituents  in  the  case  already  mentioned, 
and  in  general  in  all  cases,  show  a  molecular  weight 
(by  the  definition)  which  is  the  same  as  the  atomic  weight, 
it  is  possible  to  determine  a,  the  degree  of  ionization,  by 
osmotic-pressure  measurements,  or  from  the  average  mo- 
lecular weight  of  the  substance  in  solution,  as  determined 
by  osmotic  pressure  or  any  of  the  other  methods  mentioned 
later.  If,  for  example,  we  start  with  one  formula  weight 
of  hydrochloric  acid  in  a  solution,  and  a  moles  of  it  are 
ionized,  the  total  number  of  moles  will  consist  of  (i  —  a) 
of  un -ionized  HC1  and  a  moles  each  of  H'  and  Cl'  (where 
the  dot  indicates  positive  electricity  as  the  charge,  and 
the  accent  negative).  The  total  number  of  moles  in  the 
volume  of  the  solution  will  go  then  from  i  to  (i  —a)  +2a, 
i.e.,  1 4-  a,  and  the  ratio  of  osmotic  pressures  when  entirely  un- 
ionized, and  when  partially  ionized  will  be  the  same  as  this. 
In  other  words,  if  the  formula  weight  in  a  certain  volume 
should  give  the  osmotic  pressure  ^,  it  will  give,  when 


148  CLEMENTS   OF  PHYSICAL  CHEMISTRY. 

ionized  to  the  extent  a,  the  pressure  p  =  (i+a)pf.  Since 
the  number  of  moles  (by  definition)  shown  by  the  same 
weight  is  thus  increased,  the  molecular  weight  will  be 
smaller,  and  we  have  as  the  relation  between  the  two 
values  of  the  molecular  weight:  M(i+a)=M',  where 
the  M'-  refers  to  substance  if  it  were  un-ionized,  i.e.,  is 
the  formula  weight  of  the  substance,  and  M  is  the  average 
molecular  weight  observed. 

Just  as  with  gaseous  dissociation,  the  ionization  of 
a  solution  is  affected  by  the  presence  of  one  of  the 
products  of  the  dissociation,  and  later  when  we  consider 
the  quantitative  effect  of  this  for  gases,  we  shall  also 
consider  the  case  for  solutions. 

Owing  to  the  fact  that  the  dissociated  constituents 
in  a  solution  were  called  ions  (in  the  Faraday  sense  of 
charged  atoms),  it  has  become  common  to  speak  of  and 
to  represent  the  ionized  state  as  though  it  were  character- 
ized by  an  atomic  structure ;  in  other  words,  as  if  it  existed 
as  an  accumulation  of  charged  particles  of  infinitesimal 
size.  As  this  is  pure  hypothesis,  and  cannot  be  either 
proven  or  disproven,  we  shall  only  use  the  word  ion  here  in 

the  sense  that  it  is  expressive  of  the  relation  — ,  i.e.,  we  shall 

/*00 

only  use  it  in  the  sense  oj  ionized  matter,  according  to  this 
definition,  and  as  purely  expressive  of  an  experimentally 
determined  fact. 

One  fact  may  be  mentioned  here  which  indicates  what 


SOLUTIONS.  *49 

a  very  marked  difference  dilution  makes  in  the  behavior 
of  a  substance,  and  which  decidedly  supports  the  con- 
clusions we  have  just  drawn.  Although  hydrochloric 
acid  is  more  volatile  than  hydrocyanic  acid,  it  has  been 
observed  that  from  a  mixture  of  the  dilute  acids  (o.i 
molar  of  HC1)  it  is  possible  to  distil  the  HCN  quantitatively 
(provided  the  dilution  of  the  HC1  is  retained  at  about 
this  value  by  the  frequent  replacement  of  the  water  lost). 
In  the  light  of  the  above  theory  the  difference  between 

the  two  acids  in  solution  is  that  while  — -  is  nearly  equal 

ft* 

to  i  for  HC1,  it  is  very  small  for  HCN.  In  other  words, 
HC1  is  cofnposed  principally  of  the  ionized  constituents 
H*  and  Cl',  which  cannot  produce  HC1  gas  without 
going  through  the  state  HC1  in  solution,  and  that  is  pre- 
vented by  the  nearly  constant  dilution  which  'is  re- 
tained during  the  distillation.  Any  gaseous  substance 
then,  which  in  solution  is  largely  ionized,  is  more  difficult 
to  distil  than  an  un-ionized  or  less  ionized  one.  The 
HCN,  being  dissolved  and  retained  in  this  state  in  solution, 
can  be  expelled  readily  just  as  any  other  gas  which 
undergoes  no  great  change  in  solution.  This  method, 
indeed,  was  discovered  as  the  result  of  such  theoretical 
reasoning,  and  it  is  but  one  example  of  the  many  practical 
applications  of  the  above  generalization.  (For  details 
of  the  separation  see  Richards  and  Singer,  Am.  Chem. 
J.,  27,  205,  1902). 


15°  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

It  is  not  to  be  imagined  that  the  facts  mentioned  above 
are  the  only  ones  leading  to  these  conclusions,  for  later, 
throughout  our  work,  we  shall  find  occasion  to  consider 
other  things  which  will  confirm  each  of  the  steps  leading 
to  the  final  conclusion.  In  other  words,  it  is  not  to  be 
thought  that  the  whole  theory  has  been  described  in  this 
place,  or  that,  because  some  of  the  points  mentioned  are 
not  clear,  the  theory  itself  is  to  be  condemned,  for  many 
of  the  points  can  only  be  brought  out  after  considering 
certain  other  methods  which  will  enlarge  our  horizon. 
It  may  be  said,  however,  that  these  further  aids  but 
confirm  and  make  more  evident  the  truth  of  the  con- 
clusions we  have  arrived  at.  At  the  same  time  we  must 
not  forget  that  we  have  been  speaking  of  this  subject  as 
lying  within  certain  limits,  and  so  cannot  expect  our  con- 
clusions to  hold  outside  of  them,  nor  to  condemn  them 
because  they  do  not.  The  relation  of  substances  in 
non- aqueous  solvents  to  a  certain  extent  is  different,  and 
.consequently  these  conclusions  could  not  be  expected 
to  hold.  As  a  matter  of  fact,  the  conduction  relations 
for  these  solutions  are  so  utterly  different  from  the  aqueous 
ones  that  it  would  be  impossible  to  consider  them  together 
at  all  in  the  light  of  our  present  knowledge.  All  of  these 
points  will  be  discussed  more  fully  later,  however,  and 
the  limits  stated,  within  which  our  conclusions  in  general 
will  hold.  It  is  to  be  remembered,  however,  that  simply 
because  our  theory  does  not  hold  for  solutions  in  certain 


SOLUTIONS.  151 

non-aqueous  solvents  (solutions  which  show  no  similarity 
in  behavior  to  the  aqueous  ones,  and  which  may  or  may 
not  be  solutions,  as  we  consider  them,  but  may  involve 
an  entire  rearrangement  of  the  composition  of  the  solvent, 
or  solute,  or  both),  it  should  not  be  considered  as  false  and 
of  little  use.  For  the  two  kinds  of  systems  are  so  different 
that  it  would  be  impossible  to  imagine  that  both  are 
subject  to  the  same  laws. 

The  value  for  ct,  the  degree  of  ionization  for  a  few 
electrolytes  is  given  below  for  varying  conditions.  These 
are  the  values  as  found  from  the  ratio  of  molecular  con- 
ductivities, since  that  method  is  the  most  delicate  one 
for  this  purpose  which  we  possess. 


2 

4 
8 

16 


V 

25° 

V 

60 

a 

a 

16 

0.828 

16 

0.841 

64 

0.899 

64 

0.909 

5" 

0.962 

5" 

0.964 

4°° 

HC1 

16 

0.832 

25° 
a 

64 

0.904 

2 

0.876 

5" 

0.965 

16 

°-955 

HBr 

25° 

HI 
25° 

KC1         NaCl 

V             25°              25° 

NHsa 

LiCl 

25° 

a 

a                                          a                    a 

a 

a 

0.897 

0.895                        2         0.737 



0.932 

0.926 

10     0.86      0.842 

0.852 

0.803 

0.950 

0-945 

loo     0.94      0.937 

0.94 

0.007 

0.965 

0.963 

looo      0-98      0.982 

0.979- 



10000      o  993 

16667     0.906     . 

i$2  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

For    further  information  of    this  sort  see    Rudolphi, 
Zeit.  f.  phys.,  Chem.,  17,  385  1895. 

39.  Solution  pressure. — The  pressure  which  a  solid 
exerts  in  going  into  solution  is  called  the  solution  pressure. 
This  term,  naturally,  is  expressive  of  the  fact  that  when 
the  substance  has  dissolved  it  exerts  osmotic  pressure. 
Since  the  osmotic  pressure  increases  with  the  amount 
of  substance  dissolved,  the  solution  pressure  of  a  substance 
can  be  no  greater  than  its  osmotic  pressure  in  a  saturated 
solution,  for  that  pressure  prevents  the  substance  from 
dissolving  further.  A  substance  is  less  soluble  in  its  own 
solution  than  in  pure  water,  and  it  seems  that  it  is  the 
osmotic  pressure  which  is  due  to  that  substance  which 
effects  its  solubility  principally;  at  any  rate,  no  general 
law  is  known  which  governs  the  solubility  of  a  substance 
in  solutions  of  other  substance,  i.e.,  when  the  solution 
pressure  is  counteracted  by  an  osmotic  pressure  due  to 
another  substance.  As  will  be  seen  later,  one  of  the  ionic 
products  of  a  substance  will  when  in  solution  depress 
the  solubility  of  this  substance.  The  behavior  of  a  sub- 
stance dissolving,  then,  is  very  similar  to  a  substance 
going  into  the  gaseous  state,  and  we  have  a  solution 
pressure  corresponding  to  the  vapor-pressure. 

The  work  of  separation  of  i  mole  of  a  dissociated 
substance  from  the  volume  of  solvent  which  contained  it 
is  equal  to  iRT,  analogous  to  the  work  done  by  expansion 
of  a  dissociated  gas,  where  i  is  the  sum  of  undissociated 


SOLUTIONS.  153 

and  dissociated  portions,  i.e.,  the  total  number  of  moles 
present,  (i—  a)+na,  where  n  represents  the  number  of 
moles  of  ionized  matter  formed  from  i  mole  of  the  sub- 
stance dissolved.  Since 

iRT     . 
RT  ~*> 

a,   the  degree  of  dissociation,  may  be  readily  calculated, 

for. 

i— i 
a= . 

n  —  i 

Since  in  the  example  above  the  hydrochloric  acid  was 
almost  completely  ionized,  while  the  hydrocyanic  acid 
was  not,  the  work  of  separation  from  the  solvent,  as  given 
by  this  formula,  is  obviously  greater  for  the  former  than 
the  latter,  for  the  value  of  i  is  greater.  To  make  such  a 
separation  in  general,  then,  it  is  necessary  that  the  value 
of  iRT  exceed  that  of  RT,  and  that  by  the  dilution  the 
difference  is  made  as  marked  as  possible. 

40.  Vapor-pressures  of  solutions. — It  has  been  known 
for  many  years  that  the  vapor-pressure  of  a  solution  is 
lower  than  that  of  the  pure  solvent.  Babo  (1848)  and 
Wullner  (1856)  found,  further,  that  the  depression  0}  the 
vapor- pressure  is  proportional  to  the  amount  oj  solute 
present;  and  jor  the  same  solution  the  depression  for  any 
temperature  is  the  same  fraction  oj  the  vapor- pressure  oj 
the  pure  solvent. 

If  p  is  the  vapor-pressure  of  the  pure  solvent,  pr  that 


154  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  the  solution,  and  w  the  percentage  of   solute,  then 


•  Lj 

where  k  is  the  relative  depression,  —  —  ,  for  a  i  %  solution. 

Raoult  in  1887  found  that  a  more  general  relation  is 
obtained  if  the  relative  depression  is  determined  for  i 
mole  of  substance  in  a  certain  amount  of  solvent.  By 
experiment  he  found  that  all  solutions  containing  one 
mole  of  substance  in  a  certain  weight  of  solvent  possesses 
the  same  vapor-pressure;  i.e.,  the  MOLECULAR  depression 
of  the  vapor-  pressure  is  constant  for  all  substances  in  the 
same  solvent.  Further,  he  found  that  solutions  of  equi- 
molecular  amounts  of  substance  in  equal  weights  of 
different  solvents  give  relative  depressions  of  the  vapor- 
pressure  which  are  proportional  to  the  molecular  weights 
of  the  solvents.  This  shows  that  the  vapor-pressure 
depends  upon  the  relative  number  of  moles  of  the  solute 
and  solvent. 

Raoult  summed  up  his  results  as  follows:  One  mole 
of  any  substance  dissolved  in  100  moles  of  any  solvent 
causes  a  i%  depression  oj  the  vapor-  pressure.  (Here 
the  number  of  moles  of  solvent  is  calculated  from  the 
gaseous  molecular  weight.) 

This  law  may  also  be  expressed  in  another  form: 
The  vapor-pressure  oj  a  solution  is  to  that  oj  the  pure  sol- 
vent as  the  number  of  moles  oj  the  solvent  is  to  the  iota] 


SOLUTIONS.  155 

number  of  moles  present  in  the  solution,  i.e.  ,0/  solvent  plus 
solute. 

For  very  dilute  solutions  the  number  of  moles  of  the 
solute  is  so  small  in  comparison  with  that  of  the  solvent 
that  the  ratio  becomes  one,  i.e.,  the  vapor-pressure  of 
the  solution  is  practically  the  same  as  that  of  the  solvent. 

We  thus  obtain  another  definition  of  the  molecular 
weight  of  a  substance  in  solution,  and  this  as  far  as 
we  know  agrees  perfectly  with  our  previous  one  based 
upon  osmotic  pressure.  According  to  it;  the  molec- 
ular weight  of  a  substance  in  solution  is  that  weight 
which  in  100  formula  weights  in  grams  of  the  solvent 
depresses  the  vapor-pressure  of  this  one  hundredth  of 
its  value. 

Expressing  the  above  laws  as  equations  we  obtain 


or 


where  N  is  the  number  of  moles  of  the  solvent  and  n 
is  the  number  of  moles  of  the  solute.  This  n  is  not  the 
same  as  the  one  used  in  the  formula  (i—  a)+na,  but 
is  equal  to  this  whole  expression,  i.e.,  to  the  total  num- 
ber of  moles  present.  As  the  text  will  prevent  con- 
fusion in  these  two  values,  and  as  they  are  usually  used, 
they  are  retained  here. 


I56  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

These  equations  will  only  hold  when  the  solute  is  non- 
volatile, a  condition  which  is  practically  fulfilled  when 
its  boiling-point  is  140°  above  that  of  the  solvent,  and 
when  n  represents  the  correct  number  of  moles  in  solu- 
tion. For  ionized  substances,  then,  n  represents  the 
formula  weights  dissolved,  after  they  are  multiplied  by 
the  expression  (i  —  a)+na,  or,  in  other  words,  this 
n  is  i  times  the  number  of  formula  weights  dissolved. 
The  value  of  N  here  is  the  number  of  formula  weights 
of  solvent,  i.e.,  the  weight  in  grams  divided  by  the 
molecular  weight  of  the  solvent  in  the  gaseous  state. 

For  the  determination  of  either  the  molecular  weight 
or  the  dissociation  it  is  more  convenient  to,  use  an  altered 
form  of  equation  (27),  i.e., 


according  to  which  the  depression  oj  the  vapor-  pressure 
is  to  the  vapor-  pressure  oj  the  solution  as  the  number  oj 
moles  o]  solute  is  to  the  number  oj  moles  0}  solvent. 

w 

Since  n=—,  where  w   is    the  weight  of    solute    and 
m 

W 
m  its  molecular  weight,  and  JY=-rr,   W  and  M  being 

these  terms  for  the  solvent,  we  have 

p  —  tf     wM  wM      p' 

°r      2)     "»  =        -- 


SOLUTIONS.  157 

Example. — A  solution  of  2.47  grams  of  ethyl  ben- 
zoate  in  100  grams  of  benzene  has  a  vapor-pressure  of 
742.6  mm.  of  Hg,  while  pure  benzene  has  one  of 
751.86  mm.,  both  at  80°  C.  Find  the  molecular  weight 
of  ethyl  benzoate.  2^  =  2.47,  M  =  jS,  W  =  ioo, 
pr  =742.6,  #  =  751.86,  #-^=9.26;  hence  01  =  154, 
while  from  the  chemical  formula  we  find  C6H5COOC2H5 
=  150. 

Biddle  (Am.  Chem.  J.,  29,  341,  1903)  gives  a  dif- 
ferential method  for  determining  the  difference  in  vapor- 
pressures  of  pure  solvent  and  solution  (i.e.,  p—fl)  for 
any  concentration  and  suggests  calculating  this  to  i 
mole  of  substance  in  100  grams  of  solvent,  and  using 
it  as  the  molecular  depression  of  the  vapor-pressure. 
In  this  way,  of  course,  it  is  possible  to  determine  molec- 
ular weights  of  other  substances  in  thai  solvent  by  the 
use  of  a  simple  proportion. .  Thus  we  would  find  (p  —  pf) 
for  i  mole :  (p  —  p')  found : :  i :  x,  and  x  would  represent 

w 
the  number  of  moles,  i.e.,  — ,  contained  in  100  grams  of 

the  solvent.  Knowing  the  weight  w,  it  is  thus  easy  to 
calculate  m,  the  molecular  weight. 

For  most  organic  solvents  correct  molecular  weights 
are  obtained  by  the  use  of  these  formulae,  for  in  them 
the  dissociation  is  so  small  that  it  may  be  neglected. 

When  both  solvent  and  solute  are  volatile  it  is  still 
possible  to  calculate  the  vapor-pressure  of  the  mixture 


158  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

under  certain  conditions.  The  conditions  here  are 
these:  That  no  chemical  reaction  takes  place  between 
the  solvent  and  solute,  and  that  the  molecular  weight 
in  the  liquid  state  is  the  same  as  that  in  the  gaseous 
state,  for  both  solvent  and  solute. 

Since  in  such  a  case  the  vapor-pressure  of  each  is 
depressed  by  the  other  we  can  find  the  total  vapor-pres- 
sure by  subtracting  from  the  sum  of  the  vapor-pressures 
(i.e.,  assuming  that  neither  affects  the  other)  the  sum 
of  the  effects  of  each  on  the  other.  We  have  then 


where  the  sub-numerals  distinguish  the  two  constituents, 
the  expressions   -       —  and  -         -  represent  the  molar 


fraction  of  each  in  the  mixture,  and  the  terms  within 
the  parenthesis  are  obtained  from  (27),  considering  each 
in  turn  as  the  solvent.  By  rearrangement  this  formula 

becomes 

\  n\  n2 

(31)  x^Pi 


where  the  two  terms  represent  the  partial  pressures  of 
the  constituents  in  the  mixture,  or,  in  other  terms, 

pi  and  p2f  being  the  partial  pressures  of  each  in  the 
mixture,  fi  and  p2  having  represented  the  vapor-pres- 


SOLUTIONS.  159 

sures   at   that   temperature   of   the   pure   substances   as 
measured  alone. 

Since  by  Dalton's  law  the  total  pressure  is  equal  to 
the  sum  of  the  partial  pressures,  and  these  by  definition 
(Avagadro's  law)  are  dependent  upon  the  number  of 
moles  in  the  gaseous  space,  we  have 


and  since,  by  (31)  and  (32)  P\  =P\ 


„  /    I         / 
rl\    ~Tfl2 

ind 

Calling 

#2'  ,     -       ^2 

—7- — r=£2      and  ~=c2, 

fii' +1*2  r    '  - 


where  c2  is  the  molar  fraction  of  the  one  constituent  in 
the  gaseous  phase,  and  c2  that  of  the  same  in  the  liquid 
phase,  we  have 


/    \  / 

(33)  ^=i  -- 


160  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

and 


or 


f    \ 

(34) 


which  was  first  deduced  by  Nernst  (see  Gerber,  Dissert. 
Ueber  die  Zusammensetzung  des  Dampfes  von  Fliis- 
sigkeitsgemischen.  University  of  Jena,  1894). 

One  conclusion  which  can  be  drawn  from  (34)  is  of 
great  importance,  j  If  the  composition  of  the  gaseous 
mixture  distilling  from  a  liquid  mixture  is  the  same  as 
that  of  the  latter,  the  boiling-point  and  vapor-pressure 
of  the  mixture  is  the  same  as  that  of  the  pure  solvent. 
Here  we  must  not  confuse  such  mixtures  with  those 
which  have  a  constant  boiling-point  due  to  a  maximum 
or  minimum  of  vapor-pressure,  for  those  do  not  leave 
behind  the  same  concentration  as  that  distilling  (see  pp. 

122,     123). 

Equation  (31)  only  holds  strictly  for  solutions  of  liquids 
which  act  normally.  In  case  they  do  not,  by  an  appro- 
priate change  in  the  formula,  their  behavior  may  be 
expressed  with  some  accuracy.  Since  this  change  gives 
no  general  relation  for  all  substances,  but  necessitates 
the  knowledge  of  specific  constants  for  each  we  shall  not 


SOLUTIONS.  161 

consider  it  here.  (For  details  as  to  this  see  von  Zawidzki, 
Zeit.  f.  phys.  Chem.,  35,  129,  1900.) 

An  example  of  the  application  of  the  relation  expressed 
by  (30),  (31)  and  (32)  is  given  by  a  mixture  of  benzene 
and  ethylene  chloride  at  49.099  C.  The  sub-numeral 
i  refers  throughout  to  the  benzene. 

29.79  moles  of  ethylene  chloride,  of  which  the  vapor- 
pressure  (p2)  is  236.2  mm.  are  mixed  with  70.21  of  benzene, 
pi  =  268  mm.  What  is  the  total  vapor-pressure  of  the 
mixture?  By  (31)  we  find 

TT  =  268  Xo.702 1  +  236.2  X  0.2979  =  258.42, 

the  observed  value  being  259  mm. 

Equation  (34)  may  also  be  applied  to  these  data;    we 

have 

268  —  259 
0.2979 -£2'=       25Q      Xo.7021, 

cj  =0.2735, 

where  the  observed  value  is  0.2722,  i.e.,  the  distillate 
from  the  solution  containing  29.79  molar  per  cent  contains 
but  27.35  molar  per  cent  of  this  and  72.65  of  the  other. 

41.  The  relation  between  osmotic  pressure  and  the 
depression  of  the  vapor-pressure. — The  relation  between 
these  two  pressures  can  be  shown  by  the  following  pro- 
cess: It  is  to  be  remembered,  however,  that  the  equation 
below  (41)  is  not  the  only  way,  nor  even  the  most  con- 


162 


ELEMENTS  OF  PHYSICAL   CHEMISTRY* 


venient  one,  to %  transform  osmotic  pressures  into  vapor- 
pressures,  or  vice  versa.  Since  both  osmotic  pressure 
and  vapor-pressure  are  dependent,  according  to  what 
was  given  above,  upon  the  molar  concentrations  existing 
in  the  solutions,  the  transformation  from  one  to  the  other 
is  best  made  by  the  reduction  of  one  to  concentration, 
and  the  calculation  of  this  to  the  other.  This  process 
which  we  are  now  going  to  consider,  then,  has  simply  the 
purpose  of  showing  the  theoretical  relations  existing 
between  the  two  kinds  of  pressures,  and  will,  perhaps, 
make  them  more  real  to  the  reader.  Imagine  an  appa- 
ratus in  the  form  of  Fig.  n.  The  tube  h,  which  contains 


FIG.  ii. 


a  solution,  has  at  its  lower  end  a  semipermeable  partition, 
which  allows  passage  to  the  solvent,  but  not  to  the  solute. 
This  tube  is  placed  in  the  vessel  F,  which  contains  the 


SOLUTIONS.  163 

pure  solvent.  Assume  now  that  the  Whole  apparatus 
is  covered  with  a  bell  jar  from  which  the  air  is  exhausted. 

The  two  liquids  will  be  in  equilibrium  when  the  weight 
exerted  by  the  column  hG  is  equal  to  the  osmotic  pressure 
of  the  solution.  Both  liquids  will  evaporate,  one  at  h 
and  the  other  at  G.  The  vapor-pressure  of  the  solution 
at  h  must  then  be  the  same  as  that  of  the  solvent  at  the 
same  place.  If  this  were  not  true  liquid  would  either 
condense  or  evaporate  at  h,  and  this  would  disturb  the 
equilibrium  between  the  height  of  the  column  and  the 
osmotic  pressure  in  such  a  way  that  water  would  go 
through  the  partition.  This  would  cause,  however,  a 
continuous  cycle,  from  which  we  might  obtain  work 
without  any  change  in  temperature,  i.e.,  a  perpetual 
motion;  hence  the  vapor-pressure  of  the  two  liquids 
must  be  the  same  at  h. 

The  actual  pressure  which  the  pure  solvent  has  at  h 
is  equal  to  its  vapor-pressure  minus  the  weight  of  the 
column  of  vapor  hG.  This  is,  then,  the  vapor-pressure 
of  the  solution.  Let  there  be  N  moles  of  the  solvent  and 
n  moles  of  the  solute.  The  osmotic  pressure,  i.e.,  the 
pressure  which  the  substance  would  exert  in  gas  form, 
under  the  like  conditions,  from  (9),  for  i  mole  is 


or  for  n  moles 
(35) 


1 64  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

N  moles  of  the  solvent,  however,  will  weigh  NM  grams, 
where  M  is  the  formula  weight ;  hence 


where  5  is  the  specific  gravity  of  the  solvent.     Substituting 
this  .in  (35),  we  obtain 

P    nRTs 

= 


P,  the  osmotic  pressure,  however,  is  also  equal  to  the 
weight  of  the  column  of  solution  per  square  centimeter,  i.e., 

(37)  P  =  hs, 

the  specific  gravity  of  the  solvent  being  used  because  the 
hydrostatic  pressure  depends  only  upon  the  liquid  which 
can  go  through  the  partition.  Combining  (36)  and  (37), 
we  find 

nRT 

(30J  k=    ^    ,  T.T. 

MN 

The  weight  of  the  column  of  vapor  hG  is  proportional 
to  the  difference  between  the  vapor-pressure  of  the  solvent, 
/>,  and  that  of  the  solution,  pf,  i.e., 


where  a  is  the  specific  gravity  of  the  vapor.     In  case 


SOLUTIONS.  165 

the  column  is  of  a  considerable  height  its  density  will 
depend  upon  the  pressure,  i.e., 

adh=dp, 
or 

dp 

(39)  «-5fc- 

The  density  a,  however,  is  equal  to  -^-;  hence 


Mp 

a  =T 


or  when  combined  with  (36) 

dp    Mp 
dh~RT' 

i.e., 

dpRT=Mpdh, 

or 


and    finally   by    integration   between   the  limits    pc=P 
and    h= 


^       p 

(40) 


1  66  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Combining  this  with  (38),  we  find 

nRT    RT        p_ 

MN~  M     g"  p" 
or 

p      n 


But  this  term,  log<,  —  ,  can  be  written  in  the  form 


and  this  when  developed  in  a  series  gives 

P-P2 

- 


Here,  however,  the  second  term  may  be  dropped,  since 
p  —  p'  is  small,  and  we  find 


N' 
or 


_ 

p        N+n' 

which  is  the  equation  found  experimentally  by  Raoult. 

n 
Whenever  ]y:<o-1   we  may  use,  as  before  mentioned, 

the  simpler  equation 


P-P'  _n 

N' 


SOLUTIONS.  167 

Since  the  osmotic  pressure  is  given  by 

P=hs, 
and  h  by  (40)  is  given  as 

RT.       p 

h  =  -Ml°%*?> 

we  find 

7PT*  4) 

P  =  hs  =~77S  log,  ^  gr.-cms.  (R  =84800  gr.-cms.). 

To  obtain  the  value  of  P  in  atmospheres  it  is  only 
necessary  to  divide  this  result  by  1033  gr.-cms.,  the 
pressure  of  i  atmosphere.  We  have  then 

RT  p        RT     p-f 

(41)  P  =  -  —s  \oge  —.=  --  775  —  —  atmos. 

1033^       *  p 


Example.  —  2.47  grams  of  ethyl  benzoate  in  100  grams 
of  benzene  gives  ^  =  742.6,   while  ^  =  751.86   (at  80°) 
Find     the    osmotic     pressure.     ^=84800,     7^  =  273  +  80 
=  353°,  M  =  78,  5=0.8149,  #-^.=9.26;   hence 

P=3.6  atmospheres  at  80°  C. 

This  same  result  could,  of  course,  be  obtained  by  find- 
ing the  number  of  moles  of  ethyl  benzoate  contained 

in  i  liter  of  benzene  (i.e.,   10  X  0.8149  X~^J  and  mul- 
tiplying  this   by  the  osmotic   pressure   due    to    i    mole 


1 68  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

per  liter  at  80°  C.  In  this  way,  however,  we  obtain  the 
value  3.9  atmospheres,  instead  of  3.6,  owing  to  the  fact 
that,  as  shown  on  p.  157,  the  vapor-pressure  relation 
leads  to  a  molecular  weight  of  154,  in  place  of  the  formula 
weight  of  150,  which  we  have  used  here. 

The  ratio  of  the  vapor-pressures  ,  gives  in  general 
the  relation  between  the  number  of  moles  of  solute  to 
those  of  solvent,  i.e.,  -^.  If  we  know  the  formula  weight 

of  the  solvent  in  the  gaseous  state,  it  is  then  possible 
to  rind  the  number  of  moles  of  solute  present,  for  example, 
in  i  liter. 

The  vapor-pressure  of  an  aqueous  solution  at  o°  is 
0.95  of  that  of  the  pure  solvent,  what  is  the  osmotic  pres- 
sure of  the  solution?  According  to  the  formula 

P-P'  _n 
p'       N 

(or  either  of  the  other  two  may  be  used)  we  have 

i  — .95       n 
.95        i  OOP' 
~78~ 

where  n  is  the  number  of  moles  of  solute  in  i  liter  of 
water.  The  osmotic  pressure,  then,  is 

22.4X^  =  22.4X2.78=62.27  atmos. 


SOLUTIONS.  169 

42.  Increase  of  the  boiling-point.  —  Since  the  vapor- 
pressure  of  a  solvent  is  depressed  by  the  solution  of  a 
substance  in  it,  the  boiling-point  must  also  increase,  so 
that  from  this  term  it  is  also  possible  to  define  the  mo- 
lecular weight  of  the  substance  dissolved.  We  have 
only  to  find  for  this  purpose  the  relation  between  the 
number  of  moles  of  solute  to  the  corresponding  increase 
in  the  boiling  temperature.  This  relation  can  be  found 
by  the  aid  of  the  second  principle  of  thermodynamics, 
and  this  we  shall  do  later.  First,  however,  it  will  be  well 
to  consider  the  law  as  found  empirically,  from  the  fact 
that  the  vapor-pressure  of  a  solution  is  lower  than  that 
of  the  pure  solvent. 

One  mole  of  any  substance  dissolved  in  100  grams 
of  solvent  must  always  cause  a  certain  definite  increase 
in  the  boiling  point  of  that  solvent,  since  it  causes  a 
definite  depression  of  the  vapor-pressure.  If  A  is  the 
increase  due  to  a  i%  solution,  then  MA  is  that  due  to 
i  mole  in  100  grams  of  solvent.  We  have,  then, 

K=MA, 
or 


where  K  is  the  molecular  increase  oj  the  boiling-point, 
i.e.,  that  due  to  the  solution  oj  i  mole  oj  substance  in 


17°  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

TOO  grams  o]  solvent,  which  must  be  constant  for  all  sub- 
stances in  the  same  solvent. 

If  g  grams  of  substance  are  dissolved  in  G  grams  of 
solvent  and  the  increase  of  the  boiling-point  is  J,  then 

JG 

A  = , 

ioog' 

and  we  have  for  the  molecular  weight 


This  term  K  can  be  found  for  any  solvent  by  ascer- 
taining the  increase  in  the  boiling-point  due  to  a  cer- 
tain amount  of  a  substance  whose  molecular  weight  is 
known,  and  solving  the  equation  for  K.  Thus  i  mole 
of  any  substance  dissolved  in  100  grams  of  ether  increases 
the  boiling-point  of  the  latter  2i°.i.  This  is  calculated 
from  the  solution  of  a  smaller  amount  of  substance,  so 
that  the  observed  increase  is  much  smaller  and  much 
more  accurately  determined. 

K  may  also  be  found  as  already  mentioned  by  thermo- 
dynamical  reasoning,  by  aid  of  a  cyclic  process,  or  simply 
by  aid  of  the  second  principle.  The  general  relation  is 
derived  as  follows:  Assume  in  100  grams  of  a  solvent 
whose  formula  weight  is  M  and  whose  boiling-point, 
under  atmospheric  pressure  p,  is  T,  that  there  are  n 
moles  of  solute.  Under  the  pressure  p  the  solution 


V     QC  THE 

UNIVERSITY 

SOLUTIONS.  OF 

NA 


boils  at  the  temperature  T+dT.  At  the  temperature 
T  the  vapor-pressure  of  the  solvent  is  p,  and  consequently 
at  the  temperature  T  +  dT  it  will  be  p-\-dp.  The  differ- 
ence in  vapor-pressure,  then,  between  the  solvent  and 
the  solution  at  the  temperature  T  +  dT  is  dp. 

The  variation  of  the  vapor-pressure  with  the  temperature 
of  a  liquid  has  already  been  given  (p.  72)  as 

dp      lp 


or 


But  the  relative  depression  of  the  vapor-pressure 
caused  by  the  solution  of  n  moles  of  solute  in  100  grams  of 
solvent  is  equal  to 

dp 

dp+p' 

or,  since  dp  is  small  as  compared  to  p, 

dp 
P' 

According  to  Raoult's  law  (p.   156)  this  term  -J-  is 

iyi  p  —  p'     n  100 

equal  to  j^,  i.e.,       ,     =-^,  where  N  =~^'i  f  or  ^  here 


I72  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

is  equal  to  the  former  value  p  —  p',  and  p  is  the  vapor- 
pressure  of  the  solution,  the  jf  of  that  formula.  We 
have,  then, 

dp_^_nM^  JdT_ 

p  ~N~Too  ~2T2' 

Since,  however,  /,  the  molecular  heat  of  evaporation 
of  the  solvent,  is  equal  to  Mw,  where  w  is  the  heat  for 
i  gram,  it  follows  that 

nM  _MwdT 
100  =     2T2   ' 
or 

0.02  T2 

dT  =  -     —n, 

w 

which  for  n  =  i  reduces  to 


w 

i.e.,  the  increase  in  the  boiling-point  caused  by  the  ad- 
dition of  i  mole  of  substance  to  100  grams  of  solvent. 

Some  of  the  values  of  K,  determined  in  both  ways,  for 
they  give  the  same  results,  are:  benzene,  26.70;  chloro- 
form, 36.60;  carbon  disulphide,  23.70;  water,  5.20. 

The  osmotic  pressure  may  also  be  found  directly  from 
the  increase  in  the  boiling-point  in  the  following  way, 
since 


SOLUTIONS.  173 


7/7  T*  </>       -// 

by  substituting-™  in  place  of     ^     in  (41),  we  obtain 


Rswt 


where  w  is  latent  heat  for  i  gram,  since  l=Mw,  and  /  =dT 
is  the  increase  in  the  boiling-point.  Or  the  value  of  one 
may  be  transformed  to  the  number  of  moles  of  substance 
per  liter  of  solvent,  and  the  other  calculated  from  that. 
The  apparatus  which  is  used  for  the  practical  deter- 
mination was  devised  by  Beckmann  and  is  shown  in 
Fig.  12.  This  form  of  apparatus  has  of  late  been  changed 
in  such  a  way  that  the  heating-bath  is  eliminated,  and 
tetrahedrous  formed  by  folding  platinum  foil  substi- 
tuted for  the  garnets.  (For  details  see  Ostwald-Luther, 
Physiko-chemische  Messungen.)  About  20  cc.  of  the 
solvent  is  placed  in  the  vapor-bath  B,  which  causes 
the  boiling-tube  A  to  be  heated  evenly.  The  boiling- 
tube  has  a  piece  of  platinum  wire  fused  in  its  bottom,  and 
this  is  covered  with  small  garnets  or  beads  to  prevent  all 
bumping  and  to  cause  the  boiling  to  take  place  gently. 
On  top  of  these  beads  a  weighted  amount  of  the  solvent 
is  placed  and  the  thermometer  so  fixed  that  the  bulb  is 
covered  with  the  liquid.  The  heating  is  to  be  done 
carefully  until  the  liquid  boils,  the  part  evaporating  being 
condensed  in  the  spiral  tubes.  After  a  short  time  the 
temperature  becomes  constant  and  the  reading  of  the 
thermometer  is  taken.  This  •  gives  the  boiling-point  of 


174 


ELEMENTS   OF  PHYSICAL   CHEMISTRY. 


the  solvent,  the  amount  of  which  present  is  known.  A 
weighed  amount  of  the  substance  is  now  introduced  into 
the  tube  and  the  temperature  of  boiling  again  observed. 


FIG.  12. 

This  is  the  boiling-point  of  the  solution.  The  thermom- 
eter is  one  which  was  invented  by  Beckmann  and  is 
divided  into  hundredths  of  a  degree. 


SOLUTIONS.  175 

Example. — Beckmann  found  for  a  solution  of  2.0579 
grams  of  iodine  in  30.14  grams  of  ether  an  increase  in  the 
boiling-point  of  the  ether  equal  to  o°.566.  What  is  the 
molecular  weight  of  iodine? 

£  =  2.0579,  G=  30.14,  ^=0.566,  #  =  21.10; 
hence 


Or 


2.0579 

=  2i.ioXioo — -rr~       —  =  254. 
0.566X30.14 


2.0579 

2  1.  1  :  0.566::  M  gr.  per  100  gr.  :  -  Xioo,  ^  =  254. 

I2  corresponds  to  254;  hence  in  a  solution  of  iodine  in  ether 
the  molecular  weight  is  twice  the  usual  formula  weight. 

0.02  T2 
If  in  K  =  -  --  (p.  172)  we  substitute  the  value  of   w 

I      Ap   2T2 
(p.  72),  i.e.,  ™==  We  °btam 


According  to  Innes  (Proc.  Chem.  Soc.,  18,  26-28,  1902)  this 
formula  gives  better  values  in  some  cases  than  the  regular 
one.  It  is  possible  to  use  this  in  all  cases  where  the 
latent  heat  of  evaporation  is  unknown  and  where  the 

AT 

ratio  -:—  is.   p,  here,  is  the  pressure  under  which  the  solvent 


176  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

AT 
boils,  i.e.,  for  the  temperature  for  which  -r—  is  determined, 

and  M  is  the  molecular  weight  of  the  solvent  in  the 

AT 

gaseous    state.      Since  the  value  of  -7—   can   be  more 

Ap 

easily  and  more  accurately  determined  than  the  latent 
heat,  this  form  is  exceedingly  useful  when  using  little 
known  solvents. 

//  is  always  to  be  remembered  that  these  laws  only  hold 
when  the  pure  solvent  itself  separates.  In  case  the  solute 
also  is  volatile  a  correction  may  be  employed,  however, 
which  will  lead  to  the  correct  result.  (See  Nernst, 
Zeit.  f.  phys.  Chem.,  8,  16,  1891). 

43.  Depression  of  the  freezing-point.  —  More  than 
one  hundred  years  ago  Blagden  found  by  experiment 
that  the  freezing-point  of  a  solvent  is  depressed  by  the 
addition  of  any  substance  to  it. 

Raoult  found  further  (1887)  that  I  mole  of  any  sub- 
stance dissolved  in  100  grams  oj  any  one  solvent  causes 
a  constant  depression  oj  the  freezing- point.  If  i  mole  of 
any  substance  dissolved  in  100  grams  of  any  one  sol- 
vent causes  a  depression  equal  to  K,  then  if  A  is  the 
depression  caused  by  a  i%  solution 

MA=K, 
where  M  is  the  molecular  weight  of  the  solute,  or 


M  =  iooK-, 


SOLUTIONS.  177 

If  g  grams  of  substance  are  dissolved  in  G  grams  of 
solvent,  and  the  depression  is  J,  then 


or 


where  K  may  be  found  experimentally,  as  in  the  case  of 
the  boiling-point,  when  a  substance  of  known  molecular 
weight  is  used. 

Remembering  that  K  is  the  depression  of  the  freezing- 
point  caused  by  the  addition  of  i  mole  of  solute  to  100 
grams  of  solvent,  all  results  may  be  calculated  by  the  aid 
of  a  simple  proportion  in  place  of  the  formula  above. 
Thus,-  as  on  page  175,  we  have 

i  mole  per  100  gr.  :K:  : 

x  gr.  per  100:  depression  due  to  x  gr.  per  100. 

It  is  also  possible  to  find  an  equation  by  which  the  value 
of  K  can  be  determined  for  any  solvent.  This  equa- 
tion is  derived  as  follows  :  Assume  a  very  large  amount 
of  a  solution,  containing  P%  of  solute,  in  a  cylinder 
which  is  provided  with  a  semipermeable  piston  (Fig.  10, 
p.  138).  Let  the  freezing-point  of  the  pure  solvent  be 
r,  and  its  latent  heat  of  solidification,  i.e.,  for  i  gram, 
be  w;  further,  assume  T  —  4°  to  be  the  freezing-point 


17**  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  the  solution.  Cause  an  amount  of  the  solvent  to  be 
separated  from  the  solution  at  T°  by  a  pressure  upon  the 
piston.  If  the  amount  of  solvent  separated  previously 
contained  i  mole  of  substance  in  solution,  and  if  the 
amount  of  solution  is  so  great  that  this  has  not  caused 
an  appreciable  increase  in  the  concentration,  then  the 
osmotic  pressure,  p,  will  remain  unchanged,  and  we 
shall  have  to  do  the  osmotic  work 


Now  allow  this  volume  of  solvent  to  freeze  at  T.     By 

looMiv 
this  we  obtain  —  5  —  cais.  at  Tt  since  the  weight  of  this 

volume  is  ~~p~i  where  M  is  the  molecular  weight  of  the 

solute. 

Next  reduce  the  temperature  of  both  ice  and  solution 
J°,  i.e.,  to  T  —  J°,  and  allow  the  solid  to  melt  in  the 

solution,  by  which  —  5  —  cals.  are  absorbed  at  T  —  J°. 

Finally,  raise  the  temperature  of  the  whole  system  again 
to  T. 

The  two  amounts  of  heat,  i.e.,  the  amount  liberated 
by  the  cooling  of  the  system  J°,  and  that  absorbed  by  the 
heating  J°,  cancel.  By  this  reversible  process  we  have 
done  the  work 


SOLUTIONS.  179 


and  by  it  have  transferred  —  5  —  cals.  from  T  —  A  to  T°, 

for  heat  has  been  absorbed  at  T  —  A  and  liberated  at  T. 
By  the  second  principle  of  thermodynamics  (p.  57)  we 
have,  then, 

2T         J 


P 

or 

0.02  T2    AM 

~^T~  =  ~¥' 

AM 
This  term  ~p~>  however,  is  the  molecular  depression 

of  the  freezing-point,  i.e.,  that  due  to  a  solution  of  i  mole 
in  100  grams  of  solvent.     If  we  call  this  /,  then 

0.02  r^ 

i  —  =^v. 

W 

This  value  can  also  be  determined  in  the  following 
short  way  although  the  principle  is  identical  with  the 
above  (see  Macloskie,  Science,  N.  S.,  9,  206-207,  1899). 
Since  i  mole  of  sugar  to  the  liter  of  water  gives  an  osmotic 
pressure  of  22.4  atmospheres,  the  separation  of  the  sol- 
vent from  the  solute  would  require  the  work  of  22.4 
liter  atmospheres.  Separating  the  solvent  as  ice  would 
involve  1000X80  calories.  By  the  second  law  of  thermo- 
dynamics we  have,  then,  the  relation 


i»o  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

Work  done  Lowering  of  temperature 

Heat  during  it,  in  terms  of  work  Temperature 

i.e., 

22.4  lit.  at.  K' 


80000  X.  04  1  lit.  at.     273* 


from  which  K  =  ioK'  =  ioXi>  86  =  18.6  just  as  above. 

The  value  of  K  varies  naturally  with  the  different 
solvents,  but  is  constant  for  any  one.  Some  of  the 
observed  values  are:  water,  18.9;  acetic  acid,  38.8; 
benzene,  49.0;  phenol,  75.0. 

From  the  fact  that  the  vapor-pressure  of  a  solution 
is  lower  than  that  of  the  pure  solvent  it  is  necessary 
that  the  freezing-point  of  the  solvent  be  depressed  by 
the  addition  of  any  substance.  When  the  ice  which 
is  separated  is  the  pure  solvent,  and  the  freezing-point 
is  that  temperature  at  which  both  solid  and  liquid  may 
exist  together  in  equilibrium  in  all  proportions,  the 
vapor-pressure  of  ice  and  liquid  must  be  the  same.  This 
has  already  been  proven,  and  if  it  were  not  true  a  per- 
petual motion  would  be  possible.  In  Fig.  13  ww  is 
the  vapor-pressure  curve  for  water,  ss  that  for  a  solu- 
tion, and  ii  that  for  ice.  At  the  point  t=o  ice  and  water 
have  the  same  vapor-pressure,  and  so  are  in  equilibrium. 
The  solution  and  ice,  however,  will  only  be  in  equilibrium 
at  the  temperature  corresponding  to  the  intersection 
of  the  two  curves  i.e.,  the  freezing-point  cf  the  solution 


SOLUTIONS. 


181 


must  lie  below  that  of  pure  water.  The  more  substance 
in  the  solution,  the  lower  the  vapor-pressure  will  be  and 
the  lower  the  point  of  intersection  will  lie,  i.e.,  the  lower 
the  freezing-point  will  be.  This  is  the  same  as  the 
empirical  law  already  used  by  which  the  depression  is 
proportional  to  the  amount  of  substance  dissolved. 
We  can  also  define  the  molecular  weight  of  a  sub- 


FIG.  13. 

stance  in  solution  by  the  depression  it  produces  in  the 
freezing-point  of  the  solvent,  the  definition  depending 
upon  the  nature  of  the  solvent.  Thus  the  molecular 
weight  of  any  substance  in  water  is  that  weight  in  grams 
which  will  depress  the  freezing-point  of  100  grams  of  water 
i8.°9;  or  will  depress  it  in  the  same  ratio  when  dissolved 
in  another  amount.  ,This  holds  of  course  for  the  boil- 
ing-point as  well,  when  the  proper  increase  is  inserted. 
These  laws  also  hold  for  metals  dissolved  in  mercury, 
i.e.,  for  amalgams.  Heycock  &  Neville  have  shown 


182  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

that  potassium,  sodium,  thallium,  and  zinc  in  the  metallic 
state  all  have  a  molecular  weight  equal  to  their  atomic 
weights.  And  Hoitsema  has  shown  that  hydrogen  in 
palladium  has  a  molecular  weight  equal  to  i. 

//  is  to  be  remembered,  however,  that  these  laws  only 
hold  when  pure  solvent  solidifies. 

When  water  is  used  as  a  solvent,  and  the  substance 
is  dissociated  in  it,  the  molecular  weight  found  will  be 
smaller  than  the  formula  weight.  If  oc  is  the  true  molec- 
ular weight  and  tf  is  that  found  either  by  boiling  or 
freezing-point,  then 


i.e.,  the  ratio  of  the  true  molecular  weight  to  that  found 
is  the  same  as  that  of  the  total  number  of  moles  present 
to  the  number  present  provided  no  dissociation  takes 
place.  This  follows  from  the  fact  that  the  greater  the 
number  of  moles  present  in  a  certain  volume  the  smaller 
must  be  the  molecular  weight. 

i,  the  total  number  of  moles  present  when  we  have 
started  with  one,  is  given  by  the  equation 


or 

a  «-!+*, 

for  binary  electrolytes,  where  i-a  represents  the  frac- 
tion of  the  solute  which  is  left  in  the  undissociated  state, 


SOLUTIONS.  183 

and  20.  gives  the  number  of  moles  of  ions  formed  from 
each  decomposing  mole,  a  being  the  degree  of  disso- 
ciation. The  experimental  value  of  i  is  readily  deter- 
mined from  the  true  molecular  depression  and  that 
found,  i.e.,  for  water 

molecular  depression  found 


This  term  18.9  is  the  value  of  K  for  the  solvent  used. 

An  example  of  the  method  as  used  to  determine  molec- 
ular weight  and  the  dissociation  will  make  the  calcu- 
lation clear. 

A  solution  of  0.68  1  gram  of  acetic  acid  (which  is  very 
slightly  dissociated)  in  100  grams  of  water  causes  a 
depression  of  o°.2i68.  What  is  the  molecular  weight  of 
acetic  acid? 

#  =  18.9,     g=o.68i,     G  =  ioo,     J=o°.2i68; 

hence 

0.681 

M  =  18.9X100—  -=SQ.4. 

0.2168X100 
Or 

18.9:  0.2  i68::Mgr.  per  loogr.  :o.68i  gr.  per  100,  M  =  59.4, 
while  CH3COOH=  60. 

A  0.0107  normal  solution  of  KOH  gives  a  depres- 
sion of  o°.o388.  Find  the  degree  of  dissociation. 

Since  i  mole  in  100  grams  of  water  causes  a  depres- 


I§4  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

sion  of  1 8°. 9,  i  mole  in  1000  grams  would  cause  one  of 
i°.89.     Our  molecular  depression  is  then 


0.0388  .    3.6261 

-=3.6261,     and     ^=—      —  =  1.010. 
0.0107  1.89 


a=i— 1  =  1.919-1  =0.919,  or  the  KOH  is  91.9%  dis- 
sociated. 

The  apparatus  for  the  determination  of  the  freezing- 
point  was  devised  by  Beckmann  and  is  shown  in  Fig. 
14.  In  the  freezing-tube  A  the  solvent  (15-30  cc.)  is 
introduced  and  a  freezing-mixture  (ice  and  salt)  is  placed 
in  C.  The  temperature  is  allowed  to  fall  until  the 
liquid  is  overcooled  i  or  2  degrees;  then  it  is  stirred, 
ice  forms,  and  the  thermometer  rises  to  the  true  freezing- 
point  and  remains  constant.  The  cause  of  the  rise  in 
temperature  is  the  sudden  formation  of  ice,  which  gives 
up  heat  to  the  liquid.  B  is  a  tube  which  acts  as  an  air- 
bath  and  causes  the  cooling  to  take  place  more  evenly. 
After  the  freezing-point  of  the  pure  solvent  is  deter- 
mined the  solution  is  introduced  and  the  process  repeated, 
Or  the  solution  may  be  made  in  the  tube  by  dropping  a 
weighed  amount  of  solute  into  the  known  amount  of  the 
solvent. 

The    result,  however,  will   only  be  correct  when   the 
solid  which  separates  is  the  pure  solvent. 

One  point  to  be  kept  in  mind  is  that  the  freezing- 


SOLUTIONS. 


185 


point  thus  determined  is  of  a  solution  which  is  more 
concentrated  than  the  one  started  with,  for  some  of 
the  solvent  has  been  removed  by  the  freezing.  This 


FIG.  14. 

follows  from  the  fact  that  the  freezing-point  is  that 
temperature  at  which  ice  and  solution  exist  together 
in  equilibrium.  The  concentration  of  the  solution 
due  to  the  separation  of  ice  may  be  calculated  from 
the  overcooling  (see  p.  100),  for  it  has  been  found  for 
water  that  for  each  degree  of  overcooling  12.5  grams 


1  86  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  water  separates  from  each  liter.  Each  degree  of  over- 
cooling,  then,  causes  the  solution  to  be  1/80  more  con- 
centrated than  the  original  one.  If  the  solution  is  dilute, 
we  may  find  the  freezing-point  of  the  original  solution 
by  a  proportion.  Thus  if  the  overcooling  is  i°,  the 
depression  of  the  freezing-point  o.i,  and  v  represents 
the  volume  of  solution,  we  have 


where  x  gives  us  the  depression  of  the  freezing-point 
caused  by  the  solution  of  the  weighed  amount  of  sub- 
stance in  the  volume  v  —  i/So,  i.e.,  in  the  volume  of 
solvent  present  after  the  separation  of  solid. 

The  osmotic  pressure  may  be  found  directly  from 
the  depression  of  the  freezing-point  just  as  from  the 
increase  in  the  boiling-point,  by  substituting  in  (41)  the 

value  of  —  -77—  in  terms  of  depression  of  the  freezing- 

p-p'     dp     IdT  . 
point.      The  expression       ,     =-r=—f5  is  the   general 

relation  between  vapor-pressure  and  temperature  and 
holds  for  either  the  depression  of  the  freezing-point  or  the 
increase  of  the  boiling-point,  except  that  /  is  latent  heat 
of  fusion  in  one  case  and  of  evaporation  in  the  other. 
We  have,  then,  from  (41) 

Rswt 

P  =  -       -  —  —  atmospheres, 
I033X2XT 


SOLUTIONS.  187 

where  w  is  the  heat  of  fusion,  since  ~Tf=wi  and  t  is  the 

depression  of  the  freezing-point,  T  being  the  tempera- 
ture of  the  process. 

Here,  also,  the  transformation  of  freezing-point  depres- 
sion to  osmotic  pressure,  increase  of  the  boiling-point 
or  depression  of  the  vapor-pressure  can  be  accomplished 
by  first  finding  from  the  freezing-point  depression  the 
molar  concentration  of  the  solution  and  calculating  the 
other  values  from  this,  according  to  the  appropriate  law. 

If  we  determine  the  freezing-point  of  a  solution  and 
then  add  to  it  a  substance  which  unites  with  the  one 
already  in  solution,  thus  keeping  the  number  of  moles 
the  same,  the  new  freezing-point  will  not  differ  from 
the  original  one.  This  method  can  be  used  to  find 
the  number  of  moles  of  each  substance  uniting  to  form 
the  new  one,  for  we  know  the  origmal  number  and 
the  new  depression  will  show  how  many  are  present 
after  uniting. 

If  the  ice  which  separates  contains  solid  substance, 
the  freezing-point  of  the  solution  will  be  different  from 
that  of  the  solvent  itself,  this  one,  of  course,  being  the 
point  at  which  the  solid  solution  of  ice  and  substance 
separates.  For  such  a  rase,  naturally,  the  law  of  the 
depression  of  the  freezing-point  does  not  hold. 

44.  Division  of  a  substance  between  two  non-miscible 
solvents.  Depressed  solubility. — If  a  water  solution  of 


i88  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

succinic  acid  is  shaken  with  ether,  the  acid  is  divided 
between  the  two  solvents  in  such  a  way  that  there  is 
always  a  certain  ratio  between  the  two  concentrations, 
independent  of  the  relative  amounts  of  the  two  solvents. 
Table  IV  shows  this. 

TABLE  IV. 

H2O(i).             Ether  (2).          In  H2O  (3).  In  Ether  (4).  Coefficient  (5). 

70  cc.               30  cc.                 43-4                  7-i  6  almost 

49  cc.               49  cc.                43 . 8                  7.4  6       ' ' 

28  cc.              55-5cc-            47-4                 7-9  6  exact 

Columns  3  and  4  give  the  number  of  cc.  of  a  Ba(OH)2 
'solution  which  is  necessary  to  neutralize  100  cc.  of  the 
solutions.  The  coefficient  of  partition  depends  in  absolute 
value  upon  the  temperature  and  the  dilution  of  the  water 
solution.  For  decreased  temperature  and  concentration 
the  constant  decreases. 

//  there  are  two  or  more  substances  in  solution  the  co- 
efficients of  partition  are  the  same  as  if  each  substance 
were  present  alone. 

The  solvent  behaves  here  with  respect  to  the  substance 
dissolved  just  as  it  would  to  a  gas,  i.e.,'  between  the  two 
parts  (liquid  and  solution)  there  is  a  certain  coefficient 
absorption — the  concentration  of  substance  acting  as 
the  pressure  of  the  gas,  and  the  temperature  having  the 
same  influence  in  both  cases. 

If  the  relation  of  the  molecular  weights  remains  the 
same  throughout  the  various  concentrations  employed, 
the  coefficient  of  partition  will  remain  constant.  Thus 


SOLUTIONS.  189 

/>-nitrophenol   shows  a   coefficient   of  partition  [  — )  in 

\cw  ' 

the  system  water-chloroform,^  which  increases  with  the 
temperature  and  concentration  and  indicates  that  the 
/>-nitrophenol  is  normal  in  molecular  weight  in  water  and 
is  polymerized  in  chloroform  solution.  In  dilute  solu- 
tion the  polymerization  is  very  inconsiderable,  however. 
In  few  words,  speaking  from  the  experiments,  this  means 
that  the  greater  the  amount  of  />-nitrophenol  in  the 
system  and  the  higher  the  temperature,  the  greater  is 
the  proportion  dissolved  by  the  chloroform,  and  this 
variation  has  nothing  to  do  with  the  solubility  variations 
with  the  temperature  of  the  solvents  when  used  alone. 

An  interesting  application  of  this  was  made  by  Morse 
(Zeit.  f.  phys.  Chem.,  41,  1902)  to  find  the  amount  of 
HgCU  remaining  uncombined  in  a  complex  equilibrium. 
By  finding  the  coefficient  of  partition  of  HgCU  between 
water  and  toluol  for  known  amounts  of  HgCb  and  then 
that  for  HgCl2  between  the  aqueous  complex  solution 
and  toluol  he  was  able  to  find  the  amount  of  HgCl2 
which  was  free  and  capable  of  dissolving  from  the  complex 
solution.  (See  also  60  and  75,  Chapter  VIII.) 

If  one  solvent  is  soluble  in  another  the  amount  of  the 
one  dissolved  will  depend  upon  its  state,  i.e.,  whether 
it  is  pure  or  contains  a  substance  in  solution.  Just  as 
with  solids,  we  have  for  liquids  a  certain  solution  pressure, 
j.e,,  a  force  which  causes  them  to  dissolve.  In  the  same  way 


190  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

that  the  vapor-pressure  is  depressed  by  the  addition  of  a 
substance,  so  is  the  solubility  of  one  liquid  in  another. 
This  follows  from  the  analogy  between  gases  and  sub- 
stances in  solution.  If  s  is  the  solubility  of  A,  i.e.,  the 
amount  of  A  in  the  unit  of  volume  of  B,  and  sf  is*  the 
solubility  of  the  same  after  an  addition  of  substance  to  A, 
then  in  analogy  to 


_ 

p'       N 
we  have 

s—s'     n 

~7~  =N' 

where  N  is  the  number  of  moles  of  solvent  A  and  n  the 
number  of  moles  of  solute  in  A  . 

This  was  first  deduced  by  Nernst,  who  used  the  formula 
as  a  method  for  the  determination  of  the  molecular  weight. 

If  an  excess  of  ether  is  agitated  with  water  and  the 
freezing-point  of  the  mixture  determined  it  is  found  to  be 
lower  than  that  for  pure  water.  If  the  ethei*  contains  an 
amount  of  substance  which  is  insoluble  in  water,  then  its 
solubility  in  water  will  be  decreased,  and  the  freezing- 
point  of  the  mixture  will  be  higher  than  before.  Calling 
this  difference  the  molecular  increase  of  the  freezing- 
point,  when  the  dissolved  ether  itself  contains  i  mole  of 
dissolved  substance,  we  have  the  simple  relation 


/-/" 


SOLUTIONS.  191 

where  m  is  the  molecular  weight  of  the  substance  dissolved 
in  ether,  w  is  the  weight  of  this  in  100  grams  of  ether,  /  is 
the  freezing-point  of  a  saturated  aqueous  solution  of  ether, 
/'  the  freezing-point  of  the  saturated  aqueous  solution  of 
ether  containing  w  grams  per  100  grams,  and  ^  is  this 
molecular  increase  of  the  freezing-point,  just  defined. 
The  value  of  ^  for  a  water-ether  system  (it  varies  with  each 
system)  is  3.06. 

Table  V  gives  a  few  results  of  this  method.  The 
benzene  and  naphthalene  are  the  substances  which  in 
each  case  are  dissolved  in  the  ether 

TABLE  V. 

Substance.                               w  t-f  tn  (cal.)  m  (found) 

Benzene 2.04  0.080  78  77 

5.87  0.219  78  82 

Naphthalene 3.42  0.082  128  128 

t  in  all  cases  is  equal  to-3°.85,  i.e.,  the  freezing-point  of 

• 

a  saturated  solution  of  ether  in  water.  The  solution  of 
substance  in  ether  is  here  slightly  increased  in  concentra- 
tion, owing  to  the  separation  of  the  ether  with  the  ice,  but, 
as  very  strong  ethereal  solutions  are  rarely  used,  it  has 
but  little  influence.  Other  substances  than  ether  may 
also  be  used  with  water,  the  constant  V  of  course  being 
different  for  each  one.  Other  results  and  the  calculation 
of  the  value  of  V  which  transforms  freezing-point  results 
into  those  of  solubility  must  be  sought  in  the  original 
paper.  (Nernst,  Zeit.,  f.  phys.  Chem.,  6,  16  and  573, 
1890). 


1 92  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

45.  Solid  solutions. — Solid  solutions  are  homogeneous 
solid  phases  which  can  vary  continuously  within  certain 
limits.     These  limits  are  smaller  for  solids  than  they  are 
for  liquids,  just  as  the  limits  for  these  are  much  more 
restricted  than  for  gases.     The  most  important  differences 
between  a  solid  solution  and  a  mixture  of  solids  are  as 
follows:    Work  is  necessary  to  separate  the  constituents 
of  the  solution,  and  consequently  these  do  work  in  uniting 
to  form  the  solution;   a  solution  has  a  lower  vapor-pres- 
sure than  the  pure  solvent;  and  a  solution  has  a  smaller 
solubility  in  all  liquids  than  the  pure  solvent.     Examples 
of  solid  solutions  are  given  by  mixtures  of  isomorphous 
crystals;    zeoliths,   i.e.,   natural  water  holding  silicates, 
which    lose    water    and    yet    retain    their    transparency, 
although  the  vapor-pressure  is  reduced;    hydrogen  with 
palladium;    hydrogen  with  iron;    and  mixtures  of  two 
Crystalline  substances  which  have  different  Torms,  but 
which  give  a  mix-crystal  like  that  of  the  substance  present 
in  excess.     The  reader  must  be  referred  for  further  in- 
formation on  this  subject  to  one  of  the  books  treating  it 
in  detail,  Findlay's  Phase  Rule,  for  example. 

46.  Colloidal  mixtures.* — Colloidal  mixtures,  for  con- 
venience, are  divided  into  two  classes.     Of  these  we  shall 
consider  colloidal  solutions,  the  type  of  which  is  a  solu- 

*  For  further  information  the  reader  is  referred  to  a  paper  by  Noyes 
(Jour.  Am.  Chem.  Soc.,  27,  85,  1905),  of  which  I  have  made  extensive 
use  here. 


SOLUTIONS.  193 

tion  of  gelatine,  and  colloidal  suspensions,  of  which 
arsenious  sulphide  is  a  type. 

A  colloidal  mixture  is  a  liquid  mixture  of  two  or  more 
substances  which  are  neither  separated  by  the  long  con- 
tinued action  of  gravity  nor  by  passage  through  filter- 
paper,  although  they  are  separated  by  the  passage  through 
animal  membranes  or  close  grained  porcelain. 

Colloidal  solutions. — These  as  a  rule  are  more  or  less 
viscous,  gelatinize  upon  evaporation  and  go  into  solution 
again  on  the  addition  of  water.  They  are  not  coagu- 
lated by  the  addition  of  small  amounts  of  salts,  and,  in 
common  with  colloidal  suspensions,  influence  the  vapor- 
pressure,  freezing-point,  boiling-point  of  the  solvent  very 
much  less  than  a  corresponding  amount  of  a  crystalloid. 
This  is  also  true  for  the  osmotic  pressure  of  the  solution 
formed.  It  is  evident,  then,  employing  the  above  defini- 
tions of  molecular  weight  in  solution,  that  either  their 
molecular  weights  are  very  high,  or  the  definitions  do  not 
hold  in  such  cases.  At  present  the  question  is  still  open, 
although  from  the  enormous  results  obtained  for  the 
molecular  weights,  it  would  seem  that  the  latter  conclu- 
sion is  the  correct  one.  The  osmotic  pressure  of  a  6% 
glue  solution  has  been  found  to  be  but  "a  third  of  an  at- 
mosphere; and  the  rate  of  diffusion  is  also  found  to  be 
much  smaller  than  for  crystalloids. 

The  colloids  of  this  group  include  gelatine,  agar-agar, 
unheated  albumen,  caramel,  starch,  dextrine,  etc.  Gela- 


194  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

tinized  colloids  have  been  found  to  be  permeable  to  salts 
in  solution,  and  to  offer  but  an  exceedingly  slight  hind- 
rance to  their  passage.  The  electrical  conductivity,  as 
a  result  of  this,  of  salt  solutions  is  observed  to  be  but  a 
few  per  cent  different  from  the  value  found  in  pure  water, 
a  fact  of  which  we  shall  take  advantage  in  Chapter  IX. 
These  colloids  seem  to  be  impermeable  to  one  another, 
however. 

Colloidal  suspensions. — These,  in  contrast  to  the  other 
class,  are  coagulated  by  the  presence  of  even  a  small 
quantity  of  an  electrolyte.  They  do  not  gelatinize,  as  do 
the  others,  on  cooling,  and  if  gelatinized  do  not  redissolve 
on  heating.  The  presence  of  a  gelatinizing  colloid  in 
even  a  small  quantity  prevents  the  coagulation  of  colloidal 
suspensions  by  salts.  Thus,  in  the  presence  of  gelatine 
silver  chloride  precipitated  in  solution  remains  in  a  state 
of  colloidal  suspension,  and  this  fact  has  long  been  turned 
to  account  in  the  preparing  of  emulsions  in  photographic 
work.  It  is  true,  however,  that  a  few  other  substances 
serve  this  same  purpose,  as  sugar,  glycerol  and  even  ether. 

These  suspensions  are  not  only  coagulated  by  the 
presence  of  electrolytes,  but  are  not  affected  at  all  by  the 
presence  of  non-electrolytes.  Thus  the  addition  of  hydro- 
gen sulphide  water  to  a  solution  of  mercuric  cyanide  pro- 
duces a  colloidal  suspension,  for  neither  the  hydrocyanic 
acid  formed,  nor  the  other  two  substances  are  ionized 
beyond  a  slight  extent,  i.e.,  all  show  a  small  value  for 


SOLUTIONS.  195 

«=— .      It  seems  to  be  necessary  to  always  exceed  a 

Poo 

certain   small  minimum   quantity   of   the   electrolyte   in 
order  to  produce  coagulation. 

Metallic  colloidal  suspensions  may  be  made  (Bredig) 
by  producing  an  arc  under  pure  water,  using  electrodes 
of  the  metal  in  question.  Some  colloidal  solutions  can  be 
proven  to  consist  of  small  particles  by  examination  under 
the  microscope,  although  the  number  is  not  very  large. 
One  unfailing  property  of  all  colloidal  mixtures  is  that 
they  revolve  the  plane  of  polarized  light,  and,  in  fact,  in 
general  behave  as  though  they  are  made  up  of  exceedingly 
small  particles,  although  it  is  impossible  to  prove  that 
this  is  always  the  case. 

The  effect  of  the  electric  current  upon  colloidal  mix- 
tures is  exceedingly  important,  for  it  gives  us  some  insight 
into  their  constitution.  Certain  suspensions  are  always 
carried  in  the  direction  of  the  current  through  the  solution, 
i.e.,  coagulate  around  the  cathode  (ferric  hydroxide). 
Others,  and  this  is  by  far  the  larger  class,  collect  around 
the  anode,  i.e.,  go  with  the  negative  current.  Of  the 
first  group  we  know  of  the  basic  hydroxides  (Al,  Cr,  etc.) 
and  certain  dyestuffs.  All  the  rest,  the  ordinary  suspen- 
sions as  well  as  the  colloids,  go  in  the  direction  of  the 
negative  current  and  coagulate  around  the  anode. 

When  a  colloid  is  mixed  with  another  having  a  differ- 
ent sign,  i.e.,  when  two  colloids,  which  go  in  opposite 


*96  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

directions  under  the  influence  of  the  current,  are  mixed 
we  observe  a  coagulation,  leaving  clear  solvent  above. 
This  is  the  case  with  ferric  hydroxide  (+)  and  arsenious 
sulphide  (— ).  Non-electrolytes  have  no  effect  upon 
colloidal  mixtures,  but  electrolytes  (above  a  certain 
minimum  concentration)  cause  a  coagulation,  and  this 
effect  is  apparently  proportional  to  the  valence  of  the 
ion  of  the  opposite  sign  to  the  colloid  in  question.  Since 
the  valence  determines  the  amount  of  electricity  carried 
by  ionized  matter,  it  seems  but  a  natural  consequence 
of  our  conclusions  above,  that  ionized  matter,  and  that 
only,  outside  of  the  current,  can  cause  the  coagulation 
of  these  mixtures.  From  the  relation  with  regard  to 
the  migration  of  colloids  with  the  electric  current  it  is 
evident  that  the  colloidal  particles  carry  charges  of 
electricity,  some  always  of  one  sign,  some  of  the  other, 
depending  upon  their  nature.  And,  further,  since  non- 
electrolytes  fail  to  act  as  do  electrolytes,  with  regard 
to  coagulation,  these  facts  seem  to  confirm  our  conclu- 
sions that  the  constituents  of  electrolytes  are  charged 
with  electricity.  It  is  assumed  that  the  colloidal  particles 
are  charged  with  one  kind  of  electricity  while  the  liquid 
in  which  they  exist  has  the  opposite  charge.  Anything 
which  can  disturb  this  apparent  electrical  equilibrium, 
then,  and  the  disturbance  apparently  must  be  electrical 
itself,  causes  the  colloidal  mixture  to  coagulate. 

As  to  how  the  particles  and  solvent  become  differently 


SOLUTIONS.  197 

charged  we  have  only  hypothesis  as  yet,  or  rather  hypo- 
theses, for  there  are  several,  and  nothing  definite  is 
known. 

47.  The  molecular  weight  in  solution. — Our  definitions 
of  molecular  weight  in  solution  (osmotic  pressure,  vapor- 
pressure,  freezing-point  and  boiling-point)  in  general 
lead  us  to  the  same  results  when  it  is  possible  to  equalize 
the  experimental  error  and  employ  all  methods.  We 
have  found,  however,  that  in  some  cases  the  molecular 
weight  so  determined  is  very  much  smaller  than  the 
formula  weight  (dissociation),  and  in  other  cases  several 
times  larger  than  the  formula  weight  (association). 
Further,  two  associated  liquids  (according  to  definition 
by  surface-tension)  when  mixed  often  show  a  mutual 
dissociation  of  the  associated  state.  But  these  cases, 
excepting  the  last,  are  also  found  in  connection  with 
the  molecular  weight  in  the  gaseous  gas,  as  it  is  affected 
by  temperature  changes,  etc.  It  is  not  a  question  here 
of  any  difficulty  with  our  definition  of  molecular  weight, 
for  the  various  methods  seem  to  agree  very  well,  and 
it  is  generally  conceded  that  the  various  results  express 
the  molecular  weight  as  it  changes  from  solvent  to  solvent, 
and  in  the  same  solvent  from  dilution  to  dilution. 

Certain  results  are  found  by  the  freezing-point  method 
for  both  electrolytes  and  non-electrolytes,  however,  which 
are  hard  to  follow  so  accurately  by  other  methods,  that 
are  difficult  to  explain  other  than  by  assuming  that  there 


198  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

exists  a  hydration  of  the  substance  in  solution.  The 
freezing-point  depression,  in  other  words,  is  found  to  be 
greater  than  it  should  be  from  results  calculated  from 
data  for  more  dilute  solutions.  The  effect  of  this  hydra- 
tion would  be  small  as  to  the  increase  in  the  molecular 
weight  due  to  the  addition  of  water;  it  is  the  solvent 
which  is  removed  which  would  exert  the  greater  influence 
upon  the  result,  especially  on  the  small  amount  present 
in  concentrated  solutions.  By  plotting  actually  observed 
results  and  assuming  water  to  be  removed  from  the  sol- 
vent and  to  unite  with  the  substance,  a  curve  can  be 
found  which  will  be  in  accord  with  the  law.  In  this  way 
it  is  proven  that  ij  a  hydrate  were  formed  to  the  requi- 
site extent,  the  law  would  be  followed.  Later  we  shall 
consider  this  further  (Chap.  VIII).  Outside  of  the  one" 
experimental  fact  (see  Chap.  IX)  observed  by  Morgan 
and  Kanolt  (J.  Am.  Chem.  Soc.,  26,  635,  1904)  the  for- 
mation of  compounds  of  the  solute  and  solvent  has  not 
been  actually  observed.  According  to  the  freezing-point 
method  this  hydration  has  been  assumed  to  take  place 
with  both  the  ionized  and  the  un-ionized  portion;  the 
experimental  proof  just  mentioned  refers  only  to  the 
ionized  matter  in  a  solution  of  silver  nitrate  in  a  mix- 
ture of  water  and  pyridine. 

All  the  methods  given  above  seem  to  give  explicable 
results  for  very  dilute  solutions;  the  difficulties  will  be 
discussed  later.  For  concentrated  solutions,  however, 


SOLUTIONS.  199 

it  is  quite  evident  that  the  correct  relations  have  not 
been  obtained  as  yet.  Whether  the  behavior  of  very 
concentrated  solutions  can  ever  be  brought  into  accord 
with  the  more  dilute  ones  is  a  question  which  is  still 
open.  It  is  obvious,  though,  notwithstanding  the  diffi- 
culties which  have  arisen,  that  the  theory  of  solution  is 
a  law  of  nature  holding  between  certain  limits,  and  that 
the  problems  in  this  connection  for  the  future  have'  all 
to  do  with  enlarging  these  limits.  That  this  can  only 
be  done  by  following  the  results  closely  and  avoiding  all 
hypotheses  is  an  opinion  which  is  growing  not  alone  in 
this  branch  of  science,  but  in  all  others.  To  do  this, 
however,  it  is  not  necessary  to  cast  aside  what  we  have 
learned,  thinking  it  hypothesis,  but  to  use  the  fragment 
of  a  great  law  of  nature  which  we  undoubtedly  have  in 
the  theory  of  solution  as  a  basis  for  further  development. 


CHAPTER  VII. 
THERMOCHEMISTRY. 

48.  Definition. — Thermochemistry  is  the  subject  which 
treats   o)   the   connection   between   chemical   and   thermal 
processes.   ' 

Since  almost  every  chemical  process  is  accompanied 
by  temperature  changes,  and  is  more  or  less  influenced 
by  them,  this  subject  is  one  of  great  importance. 

If  a  substance  is  changed  from  one  state  into  another 
in  such  a  way  that  all  difference  in  energy  appears  in 
one  form,  as  heat,  for  example,  then  this  amount,  when 
determined,  is  proportional  to  the  difference  of  chemical 
energy  in  the  two  states. 

49.  Applications  of   the  principle  of  the  conservation 
of  energy. — Hess  was  the  first  to  apply  this  principle  to 
thermochemistry.     In   1840  he  announced  the  law  that 
the  amount  oj  heat  generated  by  a  chemical  reaction  is 
the  same  whether  it  takes  place  all  at  once  or  in  steps.     In 
other  words,  all  transjormations  from  the  same  original 
state  to  the  same  final  state  liberate  the  same  amount  oj 
heat,  irrespective  oj  the  process  by  which  the  final  state 

200 


THERMOCHEMISTRY.  201 

is  reached.  The  heat  liberated  in  a  reaction,  then,  de- 
pends only  upon  the  final  and  initial  states.  This  was 
entirely  the  result  of  experiment  on  the  part  of  Hess, 
who  did  not  consider  it  as  self-evident.  The  truth  of 
this  principle  can  be  shown,  for  example,  by  the  equality  in 
the  heat  of  formation  of  ammonium  chloride  in  water 
where  prepared  in  two  dissimilar  ways.  We  have 

(NH3,  HC1)  =  +42100  cal. 
(NH4Claq.)  =-   3900   " 


NH3,HC1,  aq.  =  +38200  cal. 
and 

(NH3,aq.)  -+  8400  cal. 

(HC1,  aq.)  =  +  17300  " 

(NH3,aq.,HCl,aq.)  =  + 12300   " 
(NH3,  HC1,  aq.)          =  +  38000  cal. 

Upon  this  law  the  whole  subject  of  thermochemistry 
is  based,  for  by  it  it  is  possible  to  find  indirectly  the  heat 
liberated  or  absorbed  by  any  reaction.  This  is  true  even 
though  it  is  not  possible  to  carry  out  the  reaction  in 
practice.  We  have  simply  to  consider  the  process  as  a 
part  or  a  sum  of  others  which  have  been  measured;  then, 
by  the  addition  of  the  equations,  all  undesired  substances 
may  be  caused  to  disappear  until  finally  the  desired  re- 
action is  found. 

The  chemical  symbols,  as  we  have  used  them  up  to 
now,  have  represented  only  the  molecular  weights  of  the 


202  ELEMENTS  OF  PHYSICAL    CHEMISTRY. 

substances  in  question.  Now,  however,  where  we  are 
combining  with  the  chemical  reaction  the  quantity  of 
energy  absorbed  or  liberated  by  the  reaction,  the  sym- 
bols have  a  further  meaning  which  must  not  be  over- 
looked. In  addition  to  the  weight  in  grams  expressed  by 
the  form'ula  weights  of  the  substances,  the  symbols  also 
represent  the  amount  of  energy  contained  in  that  weight 
in  this  state  as  compared  to  the  energy  contained  in 
another,  the  standard,  state.  Thus  the  equation 

C  +  2(9  =  CO  '2  +  97000  cals. 

means  that  the  energy  contained  in  12  grams  of  solid 
carbon  and  32  grams  of  gaseous  oxygen  is  greater  by 
97000  cals.  than  that  contained  at  the  same  tempera- 
ture in  44  grams  of  gaseous  carbon  dioxide. 

The  small  calorie,  cal.,  as  will  be  seen,  is  so  small 
that  the  numbers  employed  are  very  large,  and,  more 
important  still,  do  not  express  well  the  experimental 
limitations  of  the  determination.  For  this  reason  the 
Ostwald  calorie  (K)  is  much  better  adapted  for  the 
purpose.  This  is  the  heat  which  is  necessary  to  raise 
the  temperature  of  i  cc.  of  H2O  from  o°  to  100°  C.,  and 
is  related  to  the  large  and  small  calories  in  approxi- 
mately the  following  way  : 


cals.  (i.e.,  i  cc.  H2O  i°  C.); 
Cal.  (i.e.,  i  kg.  of  H2O  i°  C.). 


THERMOCHEMISTRY.  203 

It  will  be  observed  here  that  one  unit  in  terms  of  the 
Ostwald  calorie  expresses  about  the  limit  of  accuracy  of 
experimental  observations,  while  of  the  other  two,  one 
leads  to  the  impression  that  the  experimental  result  is 
more  accurate  than  it  is,  the  other  that  it  is  less  accurate. 

In  addition  to  these  units  we  also  have  others  which 
are  still  more  convenient,  for  they  are  based  upon  the 
erg.  The  joule,  designated  by  j,  is  equal  to  io7  ergs,  and 
a  unit  one  thousand  times  this,  i.e.,  io10  ergs  is  desig- 
nated by  J.  We  have,  then, 

i  cal.=4.i83  j  and  i  j  =0.2391  cal.  =io7  ergs, 
or 

i  cal.  =0.004183  J  and  i  J  =  239.1  cal.  =  io10  ergs. 

In  all  reactions  we  shall  distinguish  the  state  of 
aggregation  according  to  the  system  proposed  by  Ost- 
wald, in  which  ordinary  letters  are  used  for  liquids,  heavy 
ones  for  solids,  and  italics  for  gases. 

In  considering  substances  in  solution  it  is  necessary 
to  know  the  heat  which  is  liberated  by  this  process. 
This  is  called  the  heat  of  solution.  For  different  amounts 
of  water,  of  course,  this  will  vary,  so  that  for  uniformity 
the  heat  of  solution  is  always  understood  to  be  the  heat 
(positive  or  negative)  which  is  liberated  by  the  solution 
of  i  mole  in  so  much  water  that  an  addition  of  more 
water  will  give  no  additional  heat  effect.  This  is  desig- 
nated by  the  addition  of  the  abbreviation  Aq  (aqua) 


204  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

to  the  symbol  of  the  substance.  It  is  always  to  be  re- 
membered, however,  that  an  isolated  Aq  in  an  equation  is 
not  equal  to  an  H2O,  which  refers  to  18  grams  of  water. 
The  following  example  of  the  determination,  by  the 
process  of  elimination,  of  the  heat  liberated  by  a  reaction 
is  taken  from  the  work  of  Thomsen.  His  object  was  to 
find  the  heat  of  formation  of  SeO2Aq,  i.e.,  the  heat 
generated  when  SeO2  is  formed  from  its  elements  and 
dissolved  in  a  large  amount  of  water.  He  found 

SeO2  +  2HClAq  +  2NaHSAq 


2NaOHAq  +  2HClAq  -  2NaClAq  + 

2NaOHAq  +  2H2S  =  2NaHS  Aq  +  2  50^  ; 


By  adding  these  equations,  with  the  signs  changed  as 
indicated,  it  follows  that 


In  the  same  way  we  can  find  the  heat  of  combustion 
of  carbon  in  oxygen,  a  value  which  cannot  be  directly 
measured.  Two  reactions  which  have  been  measured 
are 

and 

CO+ 

Adding  these,  after  changing  the  signs  of  the  second,  we 
obtain 


THERMOCHEMISTRY.  205 

C+  O=CO  +  2<pK, 

i.e.,  the  sum  of  the  energies  contained  in  12  grams  of  solid 
carbon  and  16  grams  of  gaseous  oxygen  is  290^  greater 
than  that  contained  in  28  grams  of  gaseous  carbon  mon- 
oxide. 

The  process  consists,  then,  in  combining  a  number 
of  measured  reactions  in  such  a  way  that  the  final  one 
is  obtained. 

50.  The  heat  of  formation.  —  If  the  heats  of  formation 
of  the  substances  from  their  elements  are  known,  then 
it  is  simpler  to  substitute  these  in  a  reaction  and  solve 
for  the  unknown  term.  This  saves  the  trouble  of  elimi- 
nating from  a  large  number  of  equations. 

If  in  a  reaction  we  imagine  all  substances  to  be  de- 
composed into  their  elements  before  the  reaction  begins, 
then  the  final  result  of  the  reaction,  after  both  sides 
are  in  the  final  state,  will  naturally  be  the  difference 
in  the  sums  of  the  heats  of  formation  on  the  two  sides. 
We  have,  then,  the  rule:  To  find  the  heat  liberated  by 
a  reaction  it  is  simply  necessary  to  subtract  the  sum  oj 
the  heats  of  formation  of  the  original  substances  from 
that  of  those  oj  the  final  substances,  the  heat  of  formation 
of  elements  being  counted  as  zero. 

Since,  in  the  reaction, 


or 


206  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

we  can  substitute  for  the  chemical  symbols  in  an  equa- 
tion the  negative  values  of  the  heats  oj  jormation  and 
solve  jor  the  unknown  term. 

Examples.  —  Find   the   change  in  heat  energy  caused 
by  the  decomposition  of  MgCI2  by  Na.     We  have 


The  heat  of  formation  of  MgCl2  =1510^",  and  that  of 
NaCl  =  1954^,  or  2NaCl  =  3908/^5  hence 

—  1510  =  -3908  +  :*;; 
#  =  2398^. 

What    is    the    heat    of   formation    of    KMnO4?    We 
have  the  reaction 


Aq  +  5SnCl2HClAq  =  5SnQ4Aq  +  2KClAq 
+  2MnQ2Aq  +  8H2O  +  3867^, 

or,  substituting  the  negative  values  of  the  heats  of  forma 
tion, 

-2^-4057  -6290  =  -7859-  2023  -2500-  5469  +  3867; 
hence 


THERMOCHEMISTRY.  207 

To  obtain  this  result  by  the  method  of  elimination  would 
require  a  large  number  of  equations  and  be  much  more 
difficult  to  carry  out. 

The  heat  of  formation  of  organic  compounds  can  be 
found  from  the  heats  of  combustion  in  oxygen.  If  the 
elements  remain  behind  in  the  uncombined  state  the 
negative  value  of  the  one  would  give  the  other.  The 
elements,  however,  unite  to  form  H2O,  CO2,  etc.,  so 
that  the  heats  of  formation  of  these  must  be  subtracted 
from  the  heat  measured  to  obtain  the  true  value  of  the 
formation  from  the  elements. 

51.  Chemical  changes  at  a  constant  volume.  —  By 
the  first  principle  of  thermodynamics,  equation  (n),  if 
the  volume  remains  constant  the  term  pdV  disappears, 
and  we  have 

dU=dQ     (constant  F). 
By  integration  this  becomes 


where    U\   refers   to   the   internal  energy  in   the  initial 
state  and  U2  to  that  in  the  final  state.     We  have  then 


i.e.,  the  energy  in  one  state  is  equal  to  that  in  the  other 
plus  the  amount  o]  heat  generated  by  the  change.    The 


208  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

chemical  symbols  which  we  use  in  such  cases  express, 
in  addition  to  the  ordinary  chemical  meaning,  the  energy 
contained  in  I  mole. 

52.  Chemical  changes  at  a  constant  pressure.  —  If 
the  volume  changes  during  the  reaction,  i.e.,  if  a  gas 
is  formed,  then  we  may  no  longer  neglect  the  term  pdV 
in  equation  (n),  for  the  external  work  of  expansion 
absorbs  heat. 

This  term  may  be  neglected  in  all  cases  where  no 
gas  is  formed,  simply  because  the  correction  is  very 
small  as  compared  to  the  heat  of  the  reaction.  Thus, 
for  example,  we  will  calculate  the  correction  to  be  used 
for  an  increase  of  volume  equal  to  i  cc.  Here  dV  =  i 
and  ^  =  1033  grams  and  pdV  =  io^  gr.-cms.,  which  is 
equal  to 


=0.02430  cal.  or  0.000243Q.K". 


42355 


If  i  mole  of  base  is  mixed  with  i  mole  of  acid  137^" 
are  generated,  and  the  volume  increase  is  equal  to  20  cc. 
Hence  the  correction  to  be  applied  here  is 

20X0.0002439^=0.004878^, 

which  is  so  small  compared  to   137^  that  it  may  be 
neglected. 
When  a  gas  is  formed  the  increase  of  volume  may 


THERMOCHEMIS  TRY.  209 

become  very  large;    consequently  it  is  necessary  in  such 
cases  to  use  the  correction.     We  have,  equation  (n), 

dU=dQ-pdV, 

or  integrated 


or 


For  i  mole  of  gas,  however, 

pV=RT  =  2T  c&ls.  or  0.02  TK\ 
hence  for  n  moles, 

Ui  =  U2  +  QP±n(o.o2TK)  ; 

i.e.,  in  any  reaction,  for  each  mole  of  gas  formed,  at  the 
absolute  temperature  T  under  constant  pressure,  the  energy 
of  the  substance  is  decreased  by  0.02  TK. 

In  the  case  of  the  absorption  oj  the  gas  the  energy  is 
increased  by  this  amount. 

Thus  if  i  mole  of  gas  is  formed  by  a  reaction  at  18°  C. 
the  amount  of  heat  used  for  its  formation  is 

0.02  X  (273°  +i8)=5.82#. 


aid          ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

And  at  any  one  temperature  this  is  independent  of  the 
pressure  p. 

Under  constant  pressure,  the  symbols,  in  addition  to 
the  chemical  significance,  represent  the  energy  plus 
the  term  pv  jor  I  mole.  This  term  pV  may  be  either 
positive  or  negative,  according  as  the  gas  is  absorbed 
or  formed. 

53.  Relation  between  results  for  constant  volume 
and  constant  pressure.  —  We  have  for  constant  volume 


and  for  constant  pressure 


hence 


In  this  way  it  is  possible  to  make  all  our  determina- 
tions for  constant  volume  and  then  calculate  the  result 
to  constant  pressure.  This  latter  is  the  more  useful 
term,  for  all  our  reactions  take  place  in  that  way  under 
atmospheric  pressure.  The  former,  however,  -is  that 
usually  determined  in  the  bomb-calorimeter.  An  ex- 
ample of  the  calculation  from  one  condition  to  the  other 
is  given  below.  Find  from  the  reaction 

H2+  O  =H2O  +  674.84#  at  18°  const,  vol., 


THERMOCHEMISTR  Y.  211 

the  heat  generated  if  the  reaction  takes  place  under  a 
constant  pressure.  By  the  reaction  i  mole  of  hydrogen 
disappears  with  1/2  of  a  mole  of  oxygen;  consequently 
the  heat  under  constant  pressure  will  be  larger  than 
that  for  constant  volume  by  the  amount  O.O2TK  for 
each  mole  which  disappears.  We  have,  then, 

1.5(0.02X291)^=8.73, 

which  is  to  be  added  to  the  result  above,  and  gives 
674.84  +  8.73=683.57^  as  the  value  for  constant  pressure 
at  i 8°  C. 

54.  Effect  of  temperature. — If  we  allow  a  reaction  to 
take  place  first  at  the  temperature  /i  and  then  at  the 
temperature  /2  the  amounts  of  heat  evolved  are  found 
to  be  different;  assume  them  to  be  Q\  and  Q2  respect- 
ively. 

Starting  with  the  constituents,  imagine  the  reaction 
taking  place  at  the  temperature  /i,  at  which  the  amount 
of  heat  Qi  is  evolved,  and  then  heated  to  the  tempera- 
ture /2-  If  c7  is  the  heat  capacity  of  the  resulting  prod- 
ucts the  amount  of  heat  necessary  for  this  rise  in  tem- 
perature is  c/(/2-/i).  The  reaction  has  now  taken 
place  and  the  temperature  is  /2- 

Starting  again  with  the  constituents  at  the  tempera- 
ture /i  assume  them  to  be  heated  to  /2  and  then  to  react, 
evolving  the  heat  Q2-  The  heat  necessary  for  this  rise 


212  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

in  temperature  is  c(t2  —  /O,  where  c  is  the  heat  capacity 
of  the  original  substances. 

We  have  started,  thus,  with  the  same  original  sub- 
stances at  the  same  temperature  and  obtained  the  same 
products,  at  the  same  final  temperature;  hence,  by  the 
law  of  the  conservation  of  energy  (pp.  41-42)  the  amounts 
of  heat  involved  must  be  the  So,me.  We  have,  then, 

Qi-c'(t2-tl)=Q2-c(t2-t1), 
i.e., 


or 


^ 


or,  for  small  changes, 


t  A 


_ 

^orig.       ^final 


The  change  in  the  heat  of  reaction  per  degree  of  tem- 
perature is  equal  to  the  difference  of  the  molecular  specific 
heats  before  and  after  the  reaction.  It  will  be  observed, 
then,  that  the  only  change  in  the  heat  of  reaction  with 
the  temperature  is  that  due  to  the  specific  heat  of  the 
substances  present. 

It  is   to   be   noted  here   that  if  cfinai   is  larger  than 


THERMOCHEMISTRY. 


Gong,  the  sign  of  -7=   will  be  negative.     This  indicates 
at 

that  an  increase  of  temperature  gives  a  decrease  in  the 
heat  of  the  reaction,  i.e.,  the  heat  varies  inversely  with 
the  temperature.  If  c^^&\  is  smaller  than  £orig  the  term 


t 
-7-  is  positive  and  Q  varies  directly  as  the  temperature, 

Example.  —  If  4  grams  of  hydrogen  at  18°  combine  with 
32  grams  of  oxygen  at  the  same  temperature  to  produce 
36  grams  of  liquid  water,  1367.1^  are  evolved.  How  much 
heat  will  be  evolved  if  the  same  masses  of  hydrogen  and 
oxygen  combine  at  the  temperature  of  200°  and  the 
product  of  the  combination  is  maintained  at  this  tem- 
perature, the  pressure  being  constant?  The  specific 
heats  are:  3.409  for  H,  0.2175  f°r  O»  x  f°r  ft'O  from 
18°  to  100°,  and  0.4805  for  H2O  between  100°  to  200°. 
The  latent  heat  of  evaporation  of  water  is  536.5,  all 
values  being  expressed  in  small  calories. 

For  liquid  water  up  to  100°  we  have  the  relation 

(4X3.409  +  32X0.2175)  -(36X1)=-^  =  -15.404  cals., 

• 

and  since  this  is  for  i°,  we  have  for  82°  (i.e.,  100-18) 
82  X(-  15.  404)  =1263  cals.  The  heat  of  forrnation  of 
liquid  water  at  100°,  then,  is 

1367.1-12.63  =  1354.47^. 


214  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

But  in  going  into  water-  vapor  at  100°  36X536.5  cals.  of 
this  is  absorbed  by  the  system,  i.e.,  hydrogen  and  oxy- 
gen forming  gaseous  water  at  100°  evolves 

1354-47  ~  I93-H  = 

For  the  reaction  between  100°  and  200°  for  gaseous 
water,  as  formed  from  gaseous  hydrogen  and  oxygen,  we 
have  the  temperature  coefficient 


(4X3.409  +  32X0.2175)  -(36X0.4805)  =—  =2.3  cals., 


and,  for  the  interval  of  100°  (i.e.,  200  —  100), 

100  X  2.3  =  230  cals.  =  2.$K. 
In  total,  then,  we  have 

1161.33  +  2.3  =  1163.63^ 
as  the  heat  of  transformation  according  to  the  reaction 


18° 


55.  The  thermal  reactions  of  electrolytes.  —  Two  salt 
solutions  which  are  so  dilute  that  the  ratio  —  (p.  146)  is 

P-CC 

equal  to  I,  do  not  evolve  or  absorb  heat  when  mixed,  pro- 
vided no  chemical  reaction  takes  place  between  them. 


THERMOCHEMISTRY.  2  1  5 

« 

This  fact  was  first  observed  by  Hess  and  has  been 
confirmed  by  all  observers  since. 

Another  experimental  fact  observed  to  hold  for  solu- 
tions of  electrolytes  is  as  follows:  When  an  acid  is 
neutralized  by  a  base,  both  being  in  so  great  a  dilution 

that  the  value  —  for  each  is  equal  to  I,  as  is  also  that 


of  the  salt  jormed,  the  heat  evolved  is  equal  to  i^K  and 
is  independent  of  the  nature  of  the  base  and  acid  used 
or  the  salt  formed,  so  long  as  this  latter  at  that  dilution 

fulfills  the  condition  —  -  =  i. 
/**> 

These  facts,  taken  in  connection  with  those  mentioned 
above  (pp.  141-145)  and  the  conclusions  arrived  at  there, 
are  not  so  startling  as  one  might  imagine  at  first  glance. 
Since  for  the  acid  and  base  we  have  the  relation 


and,  since  the  salt  is  observed  to  have  a  molecular  weight 
(by  definition)  equal  to  one-half  the  formula  weight, 
i.e.,  is  completely  ionized  according  to  all  the  possible 
methods  of  measurement,  it  is  quite  certain  that  it  is 
made  up  of  the  substances  previously  composing  the 
acid  and  base  in  the  same  state  as  that  in  which  they 
existed  in  them.  In  other  words,  expressing  the  chem- 
ical equation  in  accord  with  the  experimental  facts  above, 
we  have 


2i6  ELEMENTS   OF  PHYSICAL    CHEMISTRY, 

9 

where  n  represents  the  number  of  moles  of  water  present 
in  the  system  before  the  reaction. 

Since  the  conductivity  shows  the  constituents  of  the 
salt  (the  ions)  to  be  present  in  the  same  form  they  were 
in  originally,  the  only  portion  of  the  reaction  which 
could  possibly  involve  heat  is  the  formation  of  water 
from  ionized  hydrogen  (H")  and  ionized  hydroxyl  (OHO- 
As  we  know  that  hydrogen  and  oxygen  in  the  ionized 
state  can  exist  together  to  but  an  infinitesimal  extent 
(for  pure  water  conducts  only  very  slightly),  the  follow- 
ing conclusion  is  certainly  justified.  When  an  acid 
unites  with  a  base  (at  any  rate  in  the  condition  in  which 
we  have  assumed  them)  the  cause  of  the  reaction  is  the 
inability  of  ionized  hydrogen  to  exist  in  the  presence  of 
ionized  hydroxyl  beyond  an  exceedingly  small  amount; 
and  the  heat  of  the  neutralization  (for  this  case)  is  that 
heat  which  is  evolved  during  the  formation  of  water 
from  its  ions  in  this  way,  i.e.,  137^  for  each  mole 
of  H*  and  OH'  (by  definition)  forming  one  mole  of 
H20. 

By  a  method  which  we  shall  consider  later  (Chap. 
VIII)  it  is  possible  not  only  to  show  the  presence  of, 
but  to  calculate  accurately,  the  heat  involved  in  the 
dissociation  of  a  substance.  When  the  acid  and  salt 
are  completely  ionized,  for  example,  and  the  base  but 
slightly,  it  is  possible  to  show  just  how  much  extra  heat 
(either  positive  or  negative)  is  involved  by  the  further 


THERMOCHEMISTRY.  217 

dissociation  of  the  base.  For  the  partly  dissociated  base 
must  increase  in  dissociation  as  its  ionized  OH  is  used 
up,  since  the  more  dilute  the  solution  of  the  base  the 
greater  is  its  ionization,  up  to  a  certain  point. 

If  both  the  acid  and  base  are  but  partly  dissociated 
the  result  will  differ  still  more,  for  heat  will  be  absorbed 
or  evolved  by  the  further  dissociation  of  both  of  these. 
In  general,  we  shall  have,  then,  if  the  salt,  also,  is  not  com- 
pletely dissociated,  i.e.,  if  more  heat  is  liberated  by  its 
undissociated  product  being  formed, 


where 

a\  =  dissociation  of  acid,   w\  =its  heat  of  dissociation, 
«2=         "  "  base,  w2="     "     " 

a3=         "  "  salt,    w3="     "     " 

x  =  heat  of  association  of  i  mole  of  H*  ions  with  i  mole 
of  OH'  ions  to  form  i  mole  of  H2O  ; 

i.e.,  the  heat  generated  by  the  neutralization  of  an  acid 
by  a  base  is  equal,  for  each  mole  oj  water  formed,  to  13?K 
plus  the  product  oj  the  heat  of  dissociation  of  the  salt  into 
the  undissociated  portion  minus  the  same  products  for 
the  acid  and  base. 

Naturally  the  negative  value  of  the  heat  of  associa- 
tion of  H'  and  OH'  ions  is  the  heat  of  dissociation  of 


2i8  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

water,  i.e.,  the  heat  necessary  to  form  i  mole  of  H'  and 
i  mole  of  OH'  ions,  from  water. 

Later  we  shall  consider  this  relation  more  in  detail, 
i.e.,  after  we  have  studied  the  method  to  be  used  for 
the  measurement  of  the  heat  of  dissociation. 

It  is  obvious  from  the  above  that  the  thermal  prop- 
erties of  electrolytes  are  additive  when  they  are  in 

such  a  dilution  that    they  fulfill  the    condition  —  =  i 

jMoo 

When  not  in  this  condition  the  change  in  the  thermal 
effect  depends  upon  the  amount  of  heat  involved  in 
causing  them  to  alter  their  states. 

When  a  precipitate  is  formed  in  such  a  solution  (i.e., 
when  a  chemical  reaction  takes  place,  which  was  excluded 
above)  it  is  often  possible  to  find  its  heat  of  formation 
just  as  we  found  that  of  water  above.  An  example  of 
this  is  the  following: 

Ag-  Aq  +  NO'3Aq  +  Na'  Aq  +  Cl'Aq  =  AgClAq  +  Na'Aq 


or 

Ag'  Aq  +  Cl'Aq  =  AgClAq  +  1  58^ 

i.e.,  when  i  mole  of  AgCl  is  formed  from  the  ionized 
silver  and  ionized  chlorine  in  a  solution  158^"  is  pro- 
duced. Conversely  if  i  mole  of  AgCl  were  dissolved; 


THERMOCHEMISTRY.  219 

this  amount  of  heat  would  be  absorbed,  i.e.,  the  heat 
of  solution  of  a  substance  is  equal  to  the  negative  value 
of  the  heat  of  precipitation. 

Although  this  is  not  always  possible,  we  can  find  the 
heat  of  formation  in  solution  in  another  way.  The  princi- 
ple of  this  is  as  follows:  By  electrical  measurements  it 
has  been  possible  to  find  the  amount  of  heat  involved 
when  2  grams  of  gaseous  hydrogen  form  2  grams  of 
ionized  hydrogen  in  solution.  This  value  is  approxi- 
mately equal  to  4  J,  but  since  there  is  some  uncertainty 
about  its  exact  value,  it  is  usual  to  assume  it  equal  to 
zero.  Later,  then,  the  results  based  upon  this  can  be 
readily  recalculated.  From  this  value,  by  dissolving  a 
metal  in  a  completely  ionized  acid,  i.e.,  by  the  substi- 
tution of  metal  in  the  ionized  state  for  the  hydrogen, 
which  is  evolved  as  a  gas  from  that  state,  we  can  observe 
directly  the  heat  of  formation  of  the  ionized  metal  from 
massive  metal.  By  then  determining  the  heat  of  solu- 
tion of  a  completely  ionized  salt  of  this  metal,  the  heat 
due  to  the  negative  radical  can  be  determined  readily, 
for  the  heat  of  solution  of  the  salt  is  equal  to  the  sum 
of  the  heats  of  ionization  of  the  constituents,  of  which 
we  assume  that  of  hydrogen  to  be  zero. 

In  this  way  the  table  given  below  has  been  prepared 
by  Ostwald.  In  order  to  find  the  heat  of  formation  of 
the  salt  it  is  only  necessary  to  form  the  sum  of  the  heats 
due  to  the  ions  into  which  it  decomposes,  taking  into 


22O 


ELEMENTS  OF  PHYSICAL    CHEMISTRY. 


account  the  valence  of  the  ions  as  indicated  by  the  dots 
for  the  electro-positive  and  the  accents  for  the  electro- 
negative substances. 


Cathions 

J  =  joules  X  io3 

Anions  of 

J  =  joules  X  io3 

Hydrogen 

H- 

+      o 

Hydrochloric  acid 

Cl' 

+    164 

Potassium 

K- 

+  259 

Hypochlorous  acid 

CIO' 

+    109 

Sodium 

Na- 

+  240 

Chloric  acid 

cio/ 

+     98 

Lithium 

Li' 

+  263 

Perchloric  acid 

CIO/ 

-    162 

Rubidium 

Rb- 

+  262 

Hydrobromic 

Br' 

+    118 

Ammonium 

NH 

4           +137 

Bromic  acid 

BrO3' 

+      47 

Hydroxylamine 

NH 

£>'  +157 

Hydriodic  acid 

I' 

+     55 

Magnesium 

Mg 

'     +456 

lodic  acid 

icv 

+   234 

Calcium 

Ca- 

+  458(?) 

Periodic  acid 

io/ 

+    i95 

Strontium 

Sr" 

+  501 

Hydrosulphuric  acid 

S" 

-     53 

Aluminium 

Al" 

+  506 

HS' 

+       5 

Manganese 
Iron 

Mn 
Fe" 

+  210 

+   93 

Thiosulphuric  acid 
Dithionic  acid 

S203" 
S200" 

+   581 
+  1166 

Fe" 

-   39 

Tetrathionic  acid 

S406" 

+  1093 

Cobalt 

Co' 

+   71 

Sulphurous  acid 

SO3" 

+   633 

Nickel 

Ni" 

+   67 

Sulphuric  acid 

SO/' 

+   897 

Zinc 

Zn" 

+  147 

Hydrogen  selenide 

Se" 

-   149 

Cadmium 

Cd- 

+   77 

Selenious  acid 

SeO3" 

+   501 

Copper 

Cu- 

-   66 

Selenic  acid 

SeO/' 

+   607 

Cu- 

-   67(?) 

Hydrogen  telluride 

Te" 

—    146 

Mercury 

Kg' 

-  85 

Tellurous  acid 

TeO3" 

+   323 

Silver 

Ag- 

-106 

Telluric  acid 

TeO/' 

+   412 

Thallium 

Tl- 

+     7 

Nitrous  acid 

NO2' 

+    H3 

Lead 

Pb- 

+        2 

Nitric  acid 

NO/ 

+   205 

Tin 

Sn" 

+    14 

Phosphorous  acid 

HPO/ 

+   603 

Phosphoric  acid 

PO/" 

+  1246 

HPO/' 

+  1277 

Arsenic  acid 

AsO/" 

+   900 

Hydroxyl 

OH' 

+   228 

Carbonic  acid 

HCO/ 

+   683 

CO/' 

+   674 

These  numbers  hold  only  for  the  case  that  the  ions  are 
in  very  dilute  solution,  i.e.,  Aq  should  be  added  to  the 
symbol  of  each.  For  stronger  solutions,  in  which  the 
ionization  is  not  complete,  other  amounts  of  heat  are 
involved  which,  unless  allowed  for,  will  lead  to  incorrect 
results. 


THERMOCHEMISTRY.  221 

The  equations 

Na  =  Na*  +  240  J 
and 


mean  that  by  the  transformation  of  the  formula  weight 
of  metallic  sodium  into  the  ionized  state  240  J  are  evolved  ; 
and  for  the  change  of  the  formula  weight  of  chlorine  gas 
into  two  formula  weights  of  ionized  chlorine  (p.  147) 
2X1647  are  liberated. 


. 

UNIVERSITY   j 

(    F 

PAL  r  rv,- 


CHAPTER  VIII. 
CHEMICAL   CHANGE. 

A.  EQUILIBRIUM. 

56.  Reversible  reactions.  —  If  we  bring  a  number 
of  reacting  substances  together  in  a  chemical  system, 
and  leave  them  for  a  sufficient  length  of  time,  the  reaction 
will  reach  an  end. 

To  represent  any  chemical  reaction  we  may  use  the 
equation 


Here  n\  moles  of  A\,  n2  of  A2)  n3  of  A3,  etc.,  unite  to 
form  n\  moles  of  A\'y  n2'  of  A2,  n3  of  A3)  etc.  When 
all  these  substances  can  remain  together  for  an  indefinite 
length  of  time,  without  the  reaction  going  in  either  direc- 
tion, they  are  said  to  exist  in  chemical  equilibrium. 

Reactions  which  go  partly  from  left  to  right  when  we 
start  with  the  substances  AI,  A2,  etc.,  and  partly  from 
right  to  left  when  we  start  with  AI,  A2  etc.,  are  called 
reversible  or  reciprocal  reactions,  provided  that  in  each 

222 


CHEMICAL   CHANGE.  223 

case,   starting  with    equivalent  amounts,   the    final  equi- 
librium is  the  same  for  both  directions. 
An  excellent  example  of  such  a  reaction  is 

C2H5OH  +  CH3COOH^±CH3COOC2H5  +  H2O. 

Alcohol.          Acetic  Acid.  Ethyl  Acetate.  Water. 

If  we  start  from  the  left  side  we  obtain  a  certain  defi- 
nite amount  of  those  on  right  and  vice  versa.  For  example, 
i  mole  (46  grams)  of  alcohol  plus  i  mole  (60  grams) 
of  acetic  acid,  or  i  mole  (88  grams)  of  ethyl  acetate 
plus  i  mole  (18  grams)  of  water,  will  always  give  the 
same  final  state,  in  which  we  have 

1/3  mole  alcohol  +  1/3  mole  acetic  acid 

+  2/3  mole  ethyl  acetate  +  2/3  mole  water. 

57.  The    law   of   mass   action.  —  Considering  such   a 
reversible  reaction  as  that  above,  or,  for  example, 


the  question  at  once  arises  —  in  which  direction  and  to 
what  extent  will  such  a  reaction  go  when  we  statf,  for 
instance,  with  a  certain  concentration  or  pressure  of 
each  of  the  three  gaseous  constituents,  HI,  I  and  H? 
From  the  purely  chemical  point  of  view  the  above 
equation  simply  provides  that  if  we  start  with  i  mole 
of  hydrogen  and  i  mole  of  iodine,  and  if  these  unite 


224  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

completely,  2  moles  of  hydriodic  acid  gas  will  be  formed; 
or  if  we  start  with  2  moles  of  hydriodic  acid  gas,  and 
this  is  completely  decomposed  we  shall  obtain  i  mole 
each  of  hydrogen  and  iodine.  As  to  what  portion  of 
the  hydrogen  and  iodine  will  unite  to  form  hydriodic 
acid;  or  what  portion  of  the  total  original  amount  of 
hydriodic  acid  will  decompose  to  form  hydrogen  and 
iodine;  or  what  will  take  place  if  all  three  are  mixed 
together;  we  are  utterly  ignorant,  failing  further  infor- 
mation than  that  contained  in  the  chemical  equation. 

The  answers  to  these  questions  can  only  be  obtained 
by  the  application  of  a  very  general  law  which  was  first 
announced  by  Guldberg  and  Waage  in  1864.  The 
qualitative  form  of  this  law  of  mass  action  is  as  follows: 
Chemical  action,  at  any  stage  of  the  process,  is  propor- 
tional to  the  active  masses  of  the  substances  present  at 
that  time,  i.e.,  to  the  amounts  of  each  present  in  the  unit 
af  volume. 

In  this  form,  however,  the  law  of  mass  action  is  of 
but  little  practical  use.  It  will  be  necessary,  then,  for 
us  to  derive  a  quantitative  expression  of  it,  and  thus 
to  obtain  it  in  such  a  form  that  it  may  be  applied  to 
our  needs  in  answering  questions  such  as  those  alluded 
to  above. 

Imagine  a  reaction  of  the  type 

n\A  i  +  n2A  $^n\A  \  +  n2'A  J 


CHEMICAL   CHANGE.  225 

having  taken  place  in  a  closed  vessel  and  to  have  attained 
a  state  of  equilibrium  in  which  we  have  the  partial  pres- 
sures Pi,  p2,  pi  and  p2r. 

Assume,  further,  that  it  is  possible  to  insert  each  of 
the  substances  on  the  left  against  its  gaseous  or  osmotic 
pressure  p\t  p2,  and  to  remove  each  of  the  products, 
as  they  are  formed,  from  the  gaseous  or  osmotic  pressure 
pi  ',  p2f  to  the  original  external  pressure  po  —  and  that 
this  insertion  and  removal  is  isothermal  and  reversible. 

Since  by  such  a  series  of  operations  we  would  do 
work  on  one  side  (—  ),  and  obtain  work  (+)  from  the 
other,  the  sum  of  the  two  amounts  (regarding  the  signs) 
would  give  us  an  expression  for  the  work  (  +  or  —  )  which 
is  done  by  the  system  itselj  during  the  transformation, 
at  constant  temperature,  of  n\  moles  of  A\  and  n2  moles 
of  A  2  to  HI  moles  of  A\  and  w2'  moles  of  A2',  the  initial 
and  final  pressure  being  the  same,  viz.,  p0.  And  this 
in  its  turn  would  lead  to  the  expression  of  the  quan- 
titative relation  existing  between  the  active  masses  of 
the  constituents  at  equilibrium,  i.e.,  to  the  relation  we 
seek. 

Since  the  work  required  to  change  the  osmotic  or 
gaseous  pressure  of  i  mole  of  substance  from  pQ  to  p\  is 

given  by  the  expression  ^riog^  —  ,*  that  for  n\  moles 

Po 


po    P  Po 


226  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

Pi 
will  be  niRTloge  —.    For  n2  moles  of  A  2  we  have  the 

'  pQ 

corresponding    expression    n2RT  log.  —  .     The    sum  of 

po 

these  two  terms  is  the  work  done,  i.e.,  lost,  by  us  in  the 
process.     The  gain  of  work  for  us  then  for  this  stage  is 


By  the  removal,  as  they  are  formed,  of  n\  moles  of  A  \ 
and  n2  moles  of  A2,  the  amount  of  work  (a  gain  for  us) 
is 


.  i 

In  total,  then,  our  gain  in  work  in  transforming  n\ 


moles  of  A  i  ^Jn<j^^t  moles  of  A2  into  n\  moles  of  A\ 
and  n2  moles  of  A2  at  constant  temperature,  the  initial 
and  final  pressure  being  p0,  is 


or 


-n2  logj>2). 


CHEMICAL   CHANGE.  227 

But,  as  we  simply  wish  to  get  the  relation  which  depends 
upon  the  pressures  in  the  reaction  at  equilibrium,  and 
the  pressure  po  has  nothing  to  do  with  this,  we  can  assume 
po  to  be  i,  and  obtain,  since  the  first  term  is  equal  to  zero, 


W=RT(nlf 


As  the  processes  of  insertion  and  removal  are  assumed 
to  be  isothermal  and  reversible  this  work,  W,  must  be 
the  maximum  work  which  can  be  done  by  the  reaction,  and 
hence  must  be  a  constant  at  any  one  temperature. 

We  have,  then, 


(42*)  W  =constant  =RT 


and  since  if  the  logarithm  is  a  constant  the  expression 
itself  must  be  a  constant,  and  since  T  and  R  are  also 
constants, 

£/Wl/  V2' 
(42)  Constant  -K  -  •*  , 


or  since  pressure  and  concentration  are  proportional, 

„  '**' 

2  K  - 


where  the  values  of  K  and  K'  may  or  may  not  be  alike, 


228  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

according  as  we  have  the  same  number  of  moles  on 
each  side  of  the  chemical  equation,  or  a  different  number. 

The  constant  (K  or  K')  is  known  as  the  constant  oj 
equilibrium. 

We  may  express  the  law  0}  mass  action  as  follows,  then : 
at  equilibrium  the  product  of  the  pressures  (concentra- 
tions) oj  the  substances  on  the  right  (final  ones),  each 
raised  to  a  power  equal  to  the  number  0}  molecules  reacting, 
divided  by  the  product  oj  the  pressures  (concentrations)  oj 
the  substances  on  the  left  (initial  ones),  each  raised  to  a 
corresponding  power,  is  a  constant  jor  any  one  reaction 
at  any  definite  temperature. 

The  variation  of  this  constant,  K  or  K'  with  the  tem- 
perature is  to  be  considered  later,  after  we  have  studied 
the  application  of  this  most  important  and  general  law. 

58.  Equilibrium  in  homogeneous  gaseous  systems. — 
For  gases  we  can  most  conveniently  use  the  form  of 
the  law  of  mass  action  which  refers  to  partial  pressures, 
(42).  We  have,  then,  for  the  equilibrium  of  a  gaseous 
chemical  system 


An  example  of  this  is  given  by  the  gaseous  reaction 


CHEMICAL   CHANGE.  229 

If  the  partial  pressure  of  H  is  pi,  that  of  Ip2y  and 
that  of  Hip,  then 


This  case  was  investigated  by  Bodenstein,  who  found 
by  experiment  that  at  a  temperature  of  boiling  sulphur 

(440°) 

- 

K  =0.02012. 


.If  we  heat  hydriodic  acid,  then,  to  this  temperature, 
it  is  possible  to  calculate  from  its  amount  the  amounts 
of  hydrogen  and  iodine  and  undecomposed  hydriodic 
acid  in  the  gaseous  state  present  at  equilibrium. 

The  total  pressure  of  a  mixture  of  gases  is  equal  to 


where  the  terms  on  the  right  are  the  partial  pressures. 
If  H  and  /  are  present  in  the  free  state  at  the  partial 
pressures  a  and  b  and  HI  to  d,  and  we  wish  to  find  in 
what  direction  and  to  -what  extent  the  reaction  will  go, 
we  proceed  as  follows:  Let  x  represent  the  partial  pres- 
sure of  H  lost;  then,  according  to  the  reaction  2HI  = 
H2  +  l2,  we  shall  have,  at  equilibrium,  for  the  partial 
pressure  of  HI 


230  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

for  H  uncombined 

PH=&-X, 

and  for  /  uncombined 


x  must  have  such  a  value,  then,  (i.e.,  the  reaction  must 
go  so  far)  ihd^^equilibrium  it  will  just  satisfy  the  equa- 
tion of  the  law  (Blmass  action,  for  the  reaction  at  440°. 


(a-x)(b-x) 

A    =O.O2OI2    = 


Knowing  a,  b,  and  d,  it  is  possible  to  solve  the  equation 
for  x,  and  to  find  how,  and  how  far  the  reaction  will 
go.  For  example,  in  this  case,  if  x  is  positive  in  value, 
the  reaction  will  go  toward  the  left,  as  we  have  assumed; 
if  negative,  in  the  opposite  direction. 

There' is  one  thing  to  be  said  of  the  solution  of  such 
equations.  There  are  two  possible  values  of  x\  which 
is  to  be  taken?  It  will  be  found  in  this  case,  as,  indeed, 
in  all  others,  that  only  one  value  is  in  accord  with  the 
existing  data,  so  that  it  alone  could  be  taken.  For 
instance,  if  the  positive  value  of  x  is  larger  than  a  or  b 
it  would  lead  to  an  absurdity,  for  it  would  show  a  nega- 
tive value  for  H  or  /,  and  the  other  value  is  the  correct 
one.  In  cases  of  equations  of  a  higher  degree,  where 


CHEMICAL   CHANGE.  231 

more  than  two  roots  exist,  this  same  rule  is  to  be  fol- 
lowed. A  possible  case  here  is  to  have  two  values  of 
the  same  sign,  but  one  smaller  than  the  other.  There 
can  be  no  question  in  such  a  case,  however,  for  if  the 
reaction  would  be  in  equilibrium  after  the  smaller  change 
had  occurred,  it  could  not  go  out  of  this  state  to  attain 
the  equilibrium  shown  by  the  greater  value,  hence  the 
lower  value  is  to  be  taken  as  the  correct  one. 

Here  we  have  used  the  partial  pressure  form  of  the 
law  of  mass  action;  we  could  use  the  other  just  as  well, 
however,  for  it  will  be  observed  that  the  constant  factor 
which  would  transform  pressures  to  concentrations 

c  =  --  •=-   is   eliminated,    since   we   have   the   same 

22.4X  - 

273 
number  of  formula  weights  (2)  on  each  side  of  the  equa- 

tion 


For  this  reason  the  constant,  K,  for  this  reaction,  as 
for  all  others  with  the  same  number  of  formula  weights 
on  both  sides,  has  the  same  value  for  concentrations, 
pressures,  volumes  under  standard  conditions,  or  any 
other  term  proportional  to  concentrations  or  pressures. 

A  further  effect  of  this  condition  of  equal  volume  on 
the  two  sides  is  that  the  progress  of  the  reaction  is  per- 
fectly independent  of  pressure  (Le  Chatelier's  theorem, 


232  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

p.  33) ;  and  Lcmoine  has  shown  this  to  be  true  for  the 
decomposition  of  HI  for  pressures  ranging  from  0.2  to 
4.5  atmospheres. 

In  using  concentrations  in  place  of  partial  pressures 
it  is  always  to  be  remembered  that  the  concentration 
(i.e.,  moles  per  liter)  is  the  actual  number  of  moles,  present 
divided  by  the  total  volume  (see  pp.  29-33).  An  exam- 
ple will  perhaps  make  this  clearer.  In  the  reaction 
A=2B  +  D,&t  equilibrium,  we  have  o.i  mole  of  A,  0.3 
of  B  and  .05  of  D  in  10  liters  at  atmospheric  pressure 
and  o°.  Starting  with  0.5  mole  of  A,  o.i  of  B  and  0.4  o£ 
D,  in  22.4  liters,  find  direction  and  extent  of  the  reaction. 

Here  we  must  first  find  the  constant  of  equilibrium 
for  the  data  given  at  equilibrium.     Since  we  have  o.i 
mole  of  A  in  10  liters,  the  concentration  of  A,  at  equilib- 
rium is  — ,  of  B  — ,  and  of  D  ^— ,  hence* 
10  10 '  10 


/o.3\2/o.o5\ 
\io/  \  io  / 


K  = 


Assuming  that  x  moles  of  A  are  formed  by  the  reac- 
tion, the  final  volume  will  be  [(o.$+x)  +  (o.i  —  2%) 
+  (0.4— #)]  22.4  liters,  the  temperature  remaining  con- 
stant at  o°,  i.e.,  (i— 2^)22.4  liters.  The  concentrations 

at  equilibrium,  then,  will  be  7 — '- — r—  -  moles  per  liter 

(i  —2^)22.4 


CHEMICAL  CHANGE.  233 

of  A,  -r—  -  r  -  of  ^>  and  7  -  of  Z>,  hence  the 

'(i  -2^)22.4  (1-200)22.4 

value  of  -K,  as   found  above,  is  to  be  equated  to  these 
values  in  the  following  way  : 


W    0.4—  x      \ 

.4/    \(l-2X)22.4/ 


Q.5+* 


r_ 


and  the  sign  of  x  will  show  the  direction  of  the  reaction, 
and  the  numerical  value  its  extent  (p.  230). 

Since  according  to  the  law  of  mass  action  the  con- 
centration is  to  be  raised  to  a  power,  it  is  the  whole  frac- 
tion representing  it  which  is  to  be  so  treated,  i.e.,  the 
number  of  moles  per  liter. 

When  applied  to  the  equilibrium  resulting  from  a 
gaseous  dissociation  the  constant  of  the  law  of  mass 
action  is  usually  designated  as  the  constant  of  dissocia- 
tion. From  it,  it  is  possible,  just  as  above,  to  calculate 
the  degree  of  dissociation  from  a  certain  amount  of  the 
dissociating  substance,  or  how  much  of  the  products, 
when  present  alone,  or  with  the  substance,  will  unite 
to  form  the  substance  itself.  And,  conversely,  we  can 
calculate  K  for  each  of  the  substances  for  which  data 
was  given  on  pages  27  and  28;  and  the  values  will  be 
dependent  only  upon  the  temperature,  the  units  employed, 
i,e.,  c  or  f,  and  nature  of  the  substance. 


234  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Where,  as  is  the  case  here,  we  know  a,  the  degree  of 
dissociation  of  the  substance,  we  can  proceed  as  follows: 
For  the  reaction 


for    example,    we    would    have    for    concentrations    the 
formula 


v 

A  = 


CPC15 


Starting  with  i  mole  of  PC/s,  which,  if  undissociated, 
would  occupy  V  liters  at  atmospheric  pressure,  with  a 
as  the  degree  of  dissociation,  the  concentrations,  where  V 
is  the  final  volume,  i.e.  (i  +  a)V  at  the  same  tempera- 
ture and  pressure,  are  as  follows  :  For  PC/5,  at  equilibrium, 

—  pr-,  for  PC/3  -y,  and   for  C/2  T?,   hence   A~,  which  we 
wish  to  determine,  is  to  be  found  from 


K 


(i-a)F  V'>'FF' 


At  250°  for  PC/5  a  =80%   (page  27)  and,  since  the 
pressure    is    atmospheric,  i    mole    must   be    present   in 

22.4       „          liters;    this  is  equal   to   V.     V,  then,  is 


CHEMICAL   CHANGE.  23 5 

equal  to  1+0.8(22.4 —  j,  and  we  have  as  the  dis- 
sociation constant  for  PCI 5  at  250° 


(i-o.8)(i+o.8)22.42732+25< 

From  the  value  thus  obtained  we  could  then  calculate 
the  direction  and  extent  of  the  reaction  at  250°  when 
we  start  with  definite  amounts  of  the  three  constituents, 
or  the  value  of  a  for  a  different  V. 

A  physical  idea  of  the  dissociation  constant,  as  found 
for  concentrations,  can  be  obtained  by  aid  of  the  formula 

a2 
\K  =  -( ^y.    Assuming  that  a  is  equal  to  0.5,  i.e.,  that 

the  degree  of  dissociation  is  50%,  for  a  reaction  by  which 

/          \2 

i  mole  is  transformed  into  2,  we  find  that  K  =  ,    \v,  or 

2K =77.    The  dissociation  constant  of  such  a  reaction,  then, 

when  multiplied  by  2  is  equal  to  the  reciprocal  oj  the  final 
volume  resulting  from  the  dissociation  of  I  mole  into  2  to  the 
extent  oj  50%.  This  volume  is  that  in  which  i  mole 
of  the  original  substance  must  be  placed  in  order  that 
at  that  temperature  it  may  dissociate  to  the  extent  of 

50%  into  two  others.  Since  77,  the  reciprocal  of  the 
volume  produced  by  the  dissociation  of  i  mole,  is  equal 


236  ELEMENTS   OF  PHYSICAL  CHEMISTRY. 

to  C,  the  concentration  in  moles  per  liter,  we  also  have 
?.K  =  C.  An  example  of  the  use  of  this  relation  is  given 
by  the  reaction  NzO^NOz  +  NOz,  for  which 

a2 
K=~.  --  ^  =  0.0138    (calculated   from  ^  =  182.69  mm., 

</d  =1.894,  dM  =  3.i8,  i.e.,  a  =0.69  and  F  =  m  all  at 
49°.  7).  Nitrogen  tetroxide,  then,  should  be  50%  disso- 

ciated at  a  concentration  of  2  X  0.0138  =  -^  =  C  moles  per 

liter;  or  at  a  dilution  of  i  mole  in  36.3  liters.  Experi- 
ment shows  that  at  this  temperature  a  =  0.493  a^  the 
dilution  i  mole  in  40  liters,  which,  considering  the  single 
value  from  which  K  is  determined,  and  the  evident 
small  error  in  observations,  is  a  satisfactory  agreement. 
A  similar  definition  could  also  be  deduced  for  the 


a 


reaction  A=2B  +  D,  where  K  =  -:—^  —  r,  although  the 


above  simpler  one  suffices  for  a  physical  idea  of  the 
dissociation  constant. 

It  is  to  be  noted  here  that  the  product  of  the  substances 
on  the  right  of  the  equation  has  always  been  placed 
in  the  numerator  of  the  fraction  giving  the  value  of  K 
(p.  227).  This  arrangement,  of  course,  is  optional,  so  long 
as  it  is  retained  the  same.  In  the  one  case  the  value  of 
K  will  simply  be  the  reciprocal  of  that  of  the  other. 

In  speaking  of  dissociation  (p.  26)  it  was  mentioned 
that  the  addition  of  one  of  the  products  of  dissociation 


CHEMICAL   CHANGE.  237 

to  the  system,  or  their  previous  presence  over  the  disso- 
ciating body,  decreases  the  extent  of  the  dissociation. 
That  this  must  be  true  according  to  the  law  of  mass 
action  is  made  obvious  by  the  consideration  of  any  defi- 
nite case. 

Supposing,   for  example,   in   the   case   of  phosphorus 
pentachloride,  the  space  over  it  contains  chlorine  prior 

cone.  C/X  cone.  PC/3 

to  the  dissociation.     Since  the  ratio  -  ~7^ri — 

cone.  PCI 5 

must  be  a  constant,  less  of  the  PCl$  will  dissociate,  for 
less  of  it,  with  the  chlorine  already  present,  will  suffice 
to  cause  the  ratio  to  attain  the  value  it  must  possess  at 
that  temperature. 

Or  to  take  another  case,  suppose  that  o.i  mole  of  B 
(p.  232)  were  introduced  into  a  vacuum,  and  the  substance 
A  allowed  to  dissociate  into  this,  arrangement  being  made 
by  a  movable  piston,  for  example,  so  that  the  final  pres- 
sure would  be  atmospheric.  What  would  be  the  effect 
of  this  o.i  mole  of  B  upon  the  dissociation  of  A  the  tem- 
perature being  o°  ?  If  the  amount  of  A  when  undisso- 
ciated  were  i  mole,  the  volume  occupied  by  it  and  thep 
o.i  mole  of  B  would  be  i.i  (22.4).  Assume  the  disso- 
ciation to  give  rise  to  oc'  moles  of  B,  then  the  number  of 
moles  of  B  in  the  final  volume  would  be  (o.i+a/),  and 

(/Y*^.  sy  ^ 

i }+-ocf,  i.e.,  i.i 
2/       2 

,  times  the  original  one,  we  havq 


238  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

_oS 
O.I+3/  ~~2~ 

•A'  A      (I.I+O22.4  ' 


and 

'  f 


K  = 


/  0.1+^   y 

\(i.  1+^)22.47  VT^T 


2 


(1.1+^)22.4 

where  the  value  of  iiT  was  found  above  (p.  232).  This 
of  is  smaller  than  the  value  (x)  which  would  be  obtained 
from  i  mole  of  A  in  the  pure  state,  occupying  the  same 
volume  at  the  same  temperature.  And  the  difference 
between  them  is  the  depression  of  the  dissociation,  in 
terms  of  B,  due  to  the  addition.  Since  for  every  mole 

x—xf 
of  A  lost,  two  of  B  are  formed,  gives  the  decrease 

of  the  dissociation  of  A  (in  moles)  due  to  the  presence 
of  the  o.i  mole  of  B. 

The  addition  of  an  indifferent  gas,  either  before  or 
after  the  dissociation,  to  a  system  composed  of  a  disso- 
ciating substance  and  its  products,  has  no  effect  upon 
the  degree  of  the  dissociation,  so  long  as  the  total  volume 
is  unchanged,  for  then  the  partial  pressures  (and  con- 
centrations) remain  unaltered.  An  increase  of  volume, 
on  the  other  hand,  no  matter  what  its  cause,  results 


CHEMICAL  CHANGE.  239 

in  an  increase  in  the  degree  of  dissociation.  When  due 
to  the  addition  of  an  indifferent  gas,  this  is  the  only 
effect,  and  the  nature  of  the  gas  is  without  influence. 
When  due  to  the  addition  of  one  of  the  constituents, 
however,  it  is  partly  compensated  by  the  depressing 
effect  of  this  upon  the  dissociation,  according  to  the 
law  of  mass  action,  and  it  is  possible  to  cause  the  one 
influence  to  just  compensate  the  other,  so  that  no  change 
in  the  dissociation  is  to  be  observed  as  the  result  of  the 
addition  with  an  increase  of  volume. 

A  somewhat  more  complicated  case  of  the  application 
of  the  law  of  mass  action  to  homogeneous  gaseous  sys- 
tems is  given  by  the  dissociation  of  carbon  dioxide,  ac- 
cording to  the  scheme, 


If  at  equilibrium  at  any  definite  temperature  the  partial 
pressures  are  p  for  CO2,  pi  for  oxygen,  and  p2  for  CO 
(where  these  come  from  the  CO2),  then  for  that  temper- 
ature 

K__PiP** 

^2     ' 

and  if  oxygen  is  already  present,  from  an  exterior  source, 
to  the  pressure  a,  the  decrease  in  the  pressure  of  carbon 
monoxide  due  to  its  effect  upon  the  dissociation  can  be 
readily  calculated.  We  have,  then, 


240  ELEMENTS  OP  PHYSICAL   CHEMISTRY. 

from  which  x  can  be  calculated  (p.  230).  p2  —  2X  will 
then  give  the  partial  pressure  of  carbon  monoxide,  and 
p  +  2x  that  of  carbon  dioxide,  in  the  presence  of  a  of. 
oxygen,  the  constant  remaining  as  above. 

For  carbon  dioxide  at  atmospheric  pressure  and  3000°, 
a  =  0.4,  i.e.,  0.5  of  the  total  pressure  is  due  to  CO2, 
0.33  to  CO,  and  0.17  to  oxygen,  consequently 

.£  =  0.074. 

The  constant  for  CO  2  may  also  have  a  different  value; 
it  is  that  which  is  obtained  from  the  formula 


and  is  equal  to 


which,  with  the  above  data,  leads  to  the  value 
£'  =  0.272. 

Naturally,  what  was  said  of  the  arrangement  of  the 
ratio  expressing  K  also  holds  here.  Either  constant  may 
be  used  for  this  temperature,  provided  that  we  always 
use  the  same  form  of  relation. 

This  is  also  true  for  the  reaction 

N20±<=±N02+N02,  or 


CHEMICAL   CHANGE. 
which  may  be  written  by  the  a  formula  cither  as 


a2  /.        a      a      i  —  a\ 

a)v   \Le"  vxv'   v  r 


or 


And  one  form  must  be  selected  and  retained. 

59.  Equilibrium  in  non-homogeneous  systems.  —  As 
we  found  above  (p.  113),  so  long  as  a  phase  of  a  system  is 
present  at  all,  its  amount  is  without  influence.  For  this 
reason  the  application  of  the  law  of  mass  action  is  simpler 
in  the  case  of  a  non-homogeneous  than  in  that  of  a  homo- 
geneous one,  for  the  active  mass  of  a  solid  phase  remains 
constant  so  long  as  it  is  present_at  all,  and  thus  can  be 
included  in  the  constant  of  equilibrium;  the  new  value 
being  a  constant,  although  not  the  one  which  would  be 
obtained  were  these  factors  regarded.  As  long  as  we 

have  the  ratio  .of  the  masses  "6f~~the  constituents  which 

*      •  *> 

regulate  the  reaction,  however/  this  constant  value  fulfils 
all  our  needs.     Thus  for  the  reaction 


calling  the  pressures  due  to  gaseous  CaCOs  and  CaO 


242  ELEMENTS  OP  PHYSICAL   CHEMISTRY. 

and  7T2,  and  that  of  the  CC>2  p->  we  have,  according  to  the 
law  of  mass  action, 


but  since  n\  and  7r2  will  remain  constant  at  any  one  tem- 
perature, we  may  employ  the  simpler  form 


i.e.,  equilibrium  at  any  one  temperature  depends  only 
upon  -the  pressure  of  the  carbon  dioxide  gas  produced. 

In   Table    VI  the  pressures   are   given   under  which 
equilibrium  exists  at  different  temperatures. 

TABLE  VI. 

EQUILIBRIUM   OF    CaCO3^±CO2+CaO. 


erature  °  C. 

Press,  mms.  of  Hg. 

Temperature  °  C.     Pi 

ess.  mms. 

547 

27 

745 

289 

610 

46 

810 

678 

625 

56 

812 

753 

740 

255 

865 

1333 

This  means  that  CaCOs  when  heated  in  a  vacuum 
to  any  temperature  gives  off  CC>2,  CaCOs,  and  CaO 
until  a  certain  pressure  is  reached.  Thus  at  547°  a 
pressure  of  27  mms.  of  Hg  is  produced. 

60.  Dissociation  of  a  solid  into  more  than  one  gas. — 
The  vapor  of  solid  NH^HS  shows  by  its  density  an  almost 
complete  dissociation  into  NH3  and  H2S, 


CHEMICAL   CHANGE.  H3 

At  2  5°.  i  the  gaseous  pressure  is  equal  to  501  mms. 
of  Hg,  i.e.,  since  the  partial  pressures  of  the  H2S  and 
NHa  are  the  same  they  are  each  equal  to  nearly  250.5 
mms.  of  Hg.  Equal  only  nearly,  however,  because  this 
pressure  includes  that  of  the  undissociated  NH4HS  gas. 
But  since  this  is  very  small  and  constant  at  any  one 
temperature  it  may  be  neglected.  If  TT  is  the  partial 
pressure  of  the  NH4HS  gas,  and  p\  and  p2  are  those 
of  the  NH3  and  H^S,  then  by  the  law  of  mass  action 


Since,  however,  TT  is  constant  at  any  one  temperature, 
we  have 


The  total  pressure,  P  =  5oi  mms.,  is  equal,  by  Dalton's 
law,  to  the  sum  of  the  partial  pressures,  and  neglecting  TT, 
we  have, 


or 


and 

PP    P* 


l2=--=- 


or 

P2 


. 

=62750. 
4          4 


244  ELEMENTS  Or  PHYSICAL   CHEMISTRY. 

This  value  K  =  pip2  may  be  verified  experimentally 
by  observing  the  effect  of  the  addition  to  the  dissoci- 
ating system  of  one  of  the  products  of  the  dissociation, 
since  the  product  of  the  partial  pressures  must  always 
remain  constant  for  constant  temperature.  Table  VII 
gives  the  results  of  experiments  carried  out  for  this 
purpose,  the  pressures  being  mms.  of  Hg. 

TABLE  VII. 


2o8  294  61152 

138  458  63204 

417  146  60882 

453  143  64779 

Average  =  62  504 

The  average  of  which  agrees  quite  well  with  the  value 
previously  found.  In  each  case  a  certain  amount  of 
one  of  the  products  is  added  before  the  solid  is  sublimed 
and  the  total  pressure  afterward  determined.  It  is 
quite  simple  then  to  find  the  amount  of  solid  which  has 
dissociated  and  thus  the  total  amount  of  each  gas  present. 
The  case  of  the  dissociation  of  ammonium  carbamate 
is  quite  similar.  We  have 


.e., 


But 


hence 


CHEMICAL   CHANGE. 


.*rP?l 


P3 
~, 


which  has  been  found  to  hold  true  by  experiment. 

The    pressure    produced    by   ammonium    carbamate 
when  heated  to  various  temperatures  is  given  below. 


tperature. 

Gaseous  Pressure. 

Temperature. 

Gaseous  Pressu, 

-15° 

2.6  mm. 

24° 

84  .  8  mm. 

-  5 

7-5 

30 

124 

o 

12.4 

40 

248 

2 

15-7 

48 

402 

4 

19.0 

5° 

470 

10 

29.8 

55 

600 

18 

53-7 

60 

770 

Isambert  has  proven  this  relation  to  hold  when  NH3 
or  CO 2  are  initially  present,  i.e.,  that  K  remains  constant 
even  after  addition  of  one  of  the  products  of  the  reaction. 
He  also  proved  that  the  presence  of  an  indifferent  gas 
does    not   affect  the   dissociation,    for   the   dissociation- 
pressure  of  a  substance  is  not  changed  by  the  presence 
of  an  indifferent  gas,  i.e.,  the  partial  pressures   remain 
unaltered.     Contrary  to  the  case  of  a  homogeneous  equi- 
librium (pp.  238  and  239),  an  increase  of  volume  has  no\ 
effect  upon  the  degree  of  dissociation  of  a  non-homogen- ! 
eous  system,  so  long  as  the  solid  (liquid)  phase  is  present,  j 
for  the  dissociation-pressure  is  dependent,  in  such  systems, 
only  upon  the  temperature  and  nature  of  the  substance. 


246  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

61.  Equilibrium  in  liquid  systems.  —  The  reaction 

CH3COOH  +  CaHsOH^CHaCOOCaHs  +  H2O, 


as  already  observed,  reaches  the  state  of  equilibrium  when 
we  have  present  1/3  mole  acid  +  1/3  mole  alcohol  +  2/3 
mole  ester  -f  2/3  mole  water,  provided  we  start  with  i  mole 
of  each  of  the  two  constituents  (either  acid  and  alcohol 
or  ester  and  water). 

This  reaction  goes  very  slowly  at  ordinary  tempera- 
tures, but  when  it  reaches  the  above  final  state  it  remains 
in  it  indefinitely.  If  we  designate  by  v  the  volume  of  the 
system,  and  start  with  i  mole  of  acid,  m  moles  of  alcohol, 
and  n  moles  of  ester  (or  water),  then  in  the  state  of  equi- 
librium, after  x  moles  of  alcohol  have  been  decomposed, 
we  shall  have 


alcohol  =  —  moles  per  liter. 


acid-— 


x  ..       .. 

ester=_  or    __ 


or     i-r     "       "     " 


CHEMICAL   CHANGE.  247 

hence,  applying  the  law  of  mass  action,  we  obtain 


A  =' 


In  the  special  case  of   equilibrium   above,   however, 
=  i,  n  =  o,  x  =  2/$;  hence 


This  value  of  K  is  one  of  the  few  which  are  practically 
independent  of  temperature.  At  10°  it  is  found  that 
65.2%  undergoes  change,  while  at  220°  the  decomposi- 
tion is  but  66.5%. 

This  equation  has  been  tested  by  experiment  with  very 
satisfactory  results.  //  has  been  found,  also,  that  by  using 
a  large  amount  of  acetic  acid  to  a  small  amount  oj  alcohol, 
or  vice  versa,  the  formation  of  ester  and  water  is  almost 
complete.  In  the  same  way  a  large .  amount  of  water 
upon  a  small  quantity  of  ester  causes  the  latter  to  be 
almost  entirely  transformed. 

In  the  following  table  some  of  the  experimental  results 
for  this  reaction  are  compared  with  the  values  calculated 
by  aid  of  the  above  formula,  and  will  serve  to  show  how 
accurate  this  law  is  in  its  application  to  liquid  systems. 


Obs. 

Calc. 

0.05 

0.049 

0.078 

0.078 

0.171 

o.  171 

0.226 

0.232 

0.293 

0.311 

0.414 

0.423 

0.519 

0.528 

0.665 

0.667 

0.819 

0.785 

0.858 

0.845 

0.876 

0.864 

0.966 

0.945 

248  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


Moles  of  Alcohol  Moles  of  Ester  or  Water, 

to  i  Mole  of  Acid. 


0.05 
O.o8 
O.l8 
0.28 

o-33 
0.50 
0.67 

I.OO 

1.50 

2.OO 
2.24 

8.00 


Using  i  mole  of  acid,  i  mole  of  alcohol,  and  various 
amounts  of  water  (no  ester  being  added)  we  find  the 
following  results. 


Moles  of  H2O  Moles  of  Ester  Formed. 

to  i  mole  of  acid 
-  1  mok  of  alcohol. 


i.o 
1-5 

2.0 

4-o 

6.5 

.1-5 


Amylene  in  contact  with  acid  forms  an  ester,  accord- 
ing to  the  equation 

\        ^ 


Obs. 

Calc. 

0.665 

0.667 

0.614 

0.596 

o-547 

0.542 

0.486 

o-S 

0.458 

0.465 

0.341 

0.368 

0.284 

0.288 

o.  198 

O.2I2 

CHEMICAL   CHANGE.  249 

If  x  is  the  amount  of  ester  formed  when  equilibrium  is 
established,  v  is  the  volume  of  the  system,  and  i  mole  of 
acid  is  used  for  a  moles  of  amylene,  then 


d  —  x 

=  amount  of  amylene  left; 


—  =      "       "  acid 


-=      "       "  ester  formed; 


hence 

xv 


The  value  for  K  'in  this  case  has  also  been  determined 
experimentally.     It  was  found  that 


K_ 

.001205* 


the  constant  in  the  form 


A  = 


having  the  value  0.001205. 


250  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

The  agreement  of  theory  and  experiment  in  this  case, 
the  acid  being  trichloracetic,  is  shown  by  the  following 
summary. 

a.  v  (liters).  ix  obs.  *  calc. 

2.15  361  0.762  0.762 

4.12  595  0.814  0.821 
4.48  638  0.820  0.826 
6.63  894  0.838  0.844 
6.80  915  0.839  0.845 

7.13  954  0.855  0.846 
7.67  1018  0.855  0.846 
9.12  1190  0-857  0.853 
9.51  1237  0.863  0.853 

14-15  *787  0.873  0.861 

As  will  be  observed,  the  volume  here,  since  there  are 
not  the  same  number  of  formula  weights  on  both  sides, 
must  be  retained  in  the  formula. 

Non-electrolytic  dissociation  in  solution. — When  a  solid 
goes  into  solution  its  action  is  apparently  analogous  to 
its  transformation  into  the  gaseous  state.  A  saturated 
solution,  thus,  in  contact  with  the  solid  at  any  tem- 
perature will  still  be  saturated.  We  have,  then,  by 
the  law  of  mass  action,  for  any  one  temperature, 

K'n=c 
or 

K  =  c, 

where  c  is  the  concentration  of  solid  in  solution  and 
varies  with  the  temperature.  If  the  solid  in  going  into 
solution  dissociates  into  other  substances,  then  an  addi- 
tion of  one  of  these  will  cause  less  substance  to  dissolve, 


CHEMICAL   CHANGE.  251 

This  has  been  proven  by  Behrend  for  a  solution  of  phenan- 
threne  picrate  in  absolute  alcohol,  in  which  a  decom- 
position into  phenanthrene  and  picric  acid  takes  place 
to  a  large  extent. 

By  the  law  of  mass  action 


where  c  =  undissociated  phenanthrene  picrate,  Ci=free 
picric  acid,  and  £2  =  free  phenanthrene  —  all  expressed 
in  moles  per  liter.  For  any  one  temperature  c  must 
be  constant,  since  the  solution  is  saturated;  hence 


=  constant. 

Coefficient  of  partition  or  distribution.  We  have  al- 
ready considered  the  distribution  of  a  substance  between 
two  non-miscible  solvents  in  the  case  that  the  formula 
weight  is  the  same  in  each  (pp.  187-189)  ;  now  we  must 
find  the  effect  of  a  difference  in  the  formula  weight.  If 
the  formula  weight  in  one  is  twice  that  in  the  other,  it  is 

A 

obvious  that  k*  €2=  c  i2,  i.e.,  that  --  and  —^  are  constant, 


for  all  dilutions.     An  illustration  of  this  is  given  by  the 
distribution  of  benzoic  acid  between  water  (cj  and  ben- 

zene (c2),  the  values  of  —  being  0.062,  0.048,  and  0.030, 
while  those  of  ~/=-,   for  the  same  dilutions,  are  0.0305, 


252  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

0.0304,  and  0.0293.  Benzoic  acid,  then,  has  twice  the 
molecular  weight  in  benzene  that  it  has  in  water,  a  fact 
which  has  been  proven  by  the  freezing-point  method. 

62.  The  effect  of  temperature  upon  an  equilibrium. 
The  variation  of  the  constant  of  equilibrium  or  dissocia- 
tion with  the  temperature.  —  In  the  case  of  an  invariant 
equilibrium  under  constant  pressure  a  change  in  the 
temperature  causes  one  of  the  phases  to  disappear 
entirely.  In  the  case  of  a  univariant  equilibrium,  how- 
ever, the  effect  is  quite  different.  A  very  small  change 
in  temperature  causes  only  a  very  slight  change  in  the 
equilibrium.  This  causes  a  corresponding  change  in 
the  relative  composition  of  the  reacting  constituents,  in 
one  direction  or  the  other,  which  just  compensates  the 
change  which  the  reaction  coefficient  has  suffered. 

By  differentiating  equation  (420)  (p.  227)  by  the  aid  of 
(42)  we  obtain 

log,  K  +  RTd(\oge  K). 


But  by  the  second  principle  (p.  58) 


where  dW  is  the  former  dQ,  i.e.,  the  work  in  terms  of  heat. 
Combining  these  two  equations,  we  get 


CHEMICAL  CHANGE.  253 

Since  W  expresses  the  work  which  is  done  by  the  trans- 
fer of  the  total  heat  Q,  the  difference,  Q-W,  is  equal 
to  the  heat  appearing  as  heat  in  the  reaction.  We  have 
thus  by  (420) 


or 


where  the  heat  of  reaction  is  given  in  terms  of  K  and  T. 

To  integrate  this  expression  it  is  necessary  to  assume 
that  q  itself  is  independent  of  the  temperature.  This 
will  undoubtedly  be  practically  true  for  small  temperature 
intervals;  for  larger  ones,  however,  we  must  be  satisfied 
to  obtain  q  as  the  value  for  the  temperature  which  is  the 
mean  of  the  two  extreme  temperatures. 

By  integration,  under  the  above  assumption,  we  find 


(43) 


Or,   using  ordinary   logarithms   and   solving  for   q,   we 
obtain 

RX  2.306  (log  K'  -log  K)TT' 

(44)        q=  T-T  ~ 


254  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

where  2.306  is  the  reciprocal  of  the  modulus  of  the  system 
of  logarithms. 

This  formula  will  then  enable  us  to  determine  the  varia- 
tion 0}  the  equilibrium  constant  K  with  the  temperature. 

Since  q  =  Q  —  W,  it  only  expresses  the  heat  of  the  re- 
action as  it  would  be  if  no  external  work  were  done  by  or 
upon  the  reaction.  The  allowance  for  this,  in  comparing 
the  results  of  (44)  with  observed  results,  must  be  made  in 
each  case,  as  is  illustrated  in  the  applications  given  below. 

One  consequence  of  this  formula  is  of  special  impor- 
tance. If  q  is  zero,  the  value  of  K  does  not  change  as  the 
result  of  a  change  in  temperature.  Thus  the  reaction 
between  acid  and  alcohol,  mentioned  above  (p.  247), 
the  mutual  transformation  of  optical  isomers,  and  a 
number  of  others,  are  found  neither  to  absorb  nor  gen- 
erate heat,  nor  to  suffer  a  displacement  of  equilibrium, 
i.e.,  a  change  in  the  value  of  K,  by  a  change  in  tempera- 
ture. 

We  shall  now  consider  the  method  of  applying  this 
equation  for  various  purposes  to  various  equilibria. 

Vaporization. — The  condition  regulating  the  equi- 
librium between  a  liquid  and  its  vapor  (a  univariant 
system)  is  the  pressure  or  concentration  of  the  latter,  and 
this  depends  upon  the  temperature.  We  have  then 


V    RT 


CHEMICAL   CHANGE.  255 

If  p  and  p  refer  to  the  two  temperatures  T  and  Tf,  then 
by  (43)>  we  nave> 


Regnault  found  for  water 

^  =  273,     ^  =  4-54  mms.  of  Hg. 
T'  =  273  +  11.54,    j/  =  10.02  mms.  of  Hg. 

From  which  by  aid  of  (44)  the  heat  of  evaporation  of 
water  at  constant  volume  is  found  for  i  mole  to  be  —  10100 
cals.  By  experiment  it  is  found  to  be  —  10854  cals.  when 

/T+T*\ 

the  volume  increases.     If  we  subtract  from  this  2  (  -  J 

=  557  cals.,  which  is  the  work  of  expansion,  we  obtain 
q=  —10297  cals. 

Dissociation  of  solids.  —  If  a  solid  dissociates  into  gases, 
the  equilibrium  is  conditioned  by  the  concentration  of  the 
latter.  If  in  the  dissociation 


ci  and  c2  are  the  concentrations  of  H  2S  and  NH3,  that 
of  the  NH4HS  being  small  and  remaining  constant,  then 


r  =-l-A  --L-JL. 

1     V    RT      2~V'~RT" 


256  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

V  meaning  in  each  case  the  volume  in  which  i  mole  is 
present. 
In  this  case 

_P_ 

2 

where  P  is  the  total  pressure;  hence 

P2 

•"-  — ^1^2  —        DTI  « 

^Kl 

If  P'  is  the  total  pressure  at  any  other  temperature  T1', 
then 

r>/2 
JTt^rJfJ^-L .. 


hence, 


or 


At  the  temperature 


=  273+  9.5,    P  =175  mms.  of  Hg; 
7  '=273  +  25.1,    P'  =  5oi  mms.  of  Hg; 


CHEMICAL   CHANGE. 

hence 

q=  -21550  cals. 

By  direct  experiment  the  molecular  heat  of  sublimation 
is  found  to  be  22800  cals.  We  must,  however,  subtract 

4  (  —      -J  cals.  from  this,  which  amount  has  been  used 
for  the  expansion.     We  have  then 

22800  —  1161  =  21639  cals.  =  —q. 

The  general  form  of  the  equation  for  this  process, 
where  n\  moles  of  one  gas  n^  of  another,  etc.,  are  given  off, 
is 


=  -- 
R\T 

This  formula  may  also  be  applied  to  the  dissociation 
of  salts  containing  water  of  crystallization  into  gaseous 
water  and  the  dehydrated  or  partially  dehydrated  salt. 

Solution  oj  solids.  —  In  this  case  the  equilibrium  depends 
only  upon  the  concentration  of  soKd  substance  in  the 
solution,  and  the  temperature;  we  have,  then, 

K=c, 

where  c  is  the  concentration  of  a  saturated  solution  at 
the  temperature   T.    If  d  is  the  concentration  of  such 


258  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

a  solution  at  another  temperature  T',  then  it  is  possible 
for  us  to  calculate  by  (44)  the  heat  of  solution  of  the 
solid,  the  increase  of  volume  being  so  small  that  it  is 
practically  equal  to  zero. 

van't  HofT  found  by  experiment  with  succinic  acid 
in  water  that 

for  T  =  273,  c  =  2.88  moles  per  liter; 

and         for  ^'  =  273  +  8.5,     c'— -4.22  moles  per  liter. 
For  i  mole,  then,  by  (44), 

q=-  —6900  cals., 

while  Berthelot  found  — 6700  cals.  by  direct  experiment. 

lonization  of  solids  in  solution. — If  a  substance  is  very 
slightly  soluble,  then  the  solution  must  contain  prin- 
cipally ions  and  very  little  undissociated  substance,  and 
the  heat  of  dissociation  must  be  the  same  as  the  heat 
of  solution,  i.e.,  equal  to  the  negative  value  of  the  heat 
of  precipitation  from  the  free  ions.  Thus  for  AgCl  we 

have 

AgCl=Ag' 


If  the  solubility  at  T  =  c,  and  at  T'  =  c?9  in  moles  per 
liter,  then,  since  2  moles  of  the  ions  form  from  i  mole  of 
the  salt,  we  have,  as  on  page  257, 


CHEMICAL   CHANGE.     .  259 

For  7^  =  273  +  20,      c  =  i.ioXio~5, 


hence  q=—i  5900  =  —  1  59^. 

For  the  negative  heat  of  precipitation  we  found  (p.  218) 
—  I5&K,  which  is  an  excellent  agreement.  We  have 
assumed  the  ionization  to  be  complete  here  and  the 
fact  that  the  heat  results  agree,  cannot  but  be  considered 

as  confirmatory  of  our  assumption. 
^^" 

Dissociation  oj  gaseous  bodies.  —  When  a  substance  A 

dissociates  according  to  the  scheme 


the  equation  of  equilibrium  is 


where  c,  c\,  c2,  .  .  .  are  the  concentrations  of  A,  A\, 
A2,  .  .  .  in  moles  per  liter. 

If  the  mole  occupies  the  volume  V  at  T  and  the 
volume  V  at  T',  then  for  the  dissociation  of  N2C>4  we 
have 


hence  (p.  241) 

a'2  a2 


/  v  ,  v 

(i—  a')V  (i—  a)V 


260  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

and 

IOP"    •  —  IOP    =  — I —  I 

lue«  /j  _  a'\y     1U&<?  (i—a)y     R\T     Tr  I 

If  we  assume  the  vapor-densities  to  be  A  and  Ar  at 
T  and  !F,  then 


,,       , 
and     «'  = 


where  3.179  is  the  vapor-density  of  N2CU  as  calculated 
from  the  molecular  weight  (96)  and  based  upon  air  as 
a  standard. 

The  volume  occupied  by  i  mole  of  N2O4  under  atmos- 
pheric pressure  is  increased  by  the  dissociation.    One  mole 

(22-4\ 
from  ~~6J  • 
/  o 

The  volume  in  which  i  mole  is  present  will  be  increased, 
then,  to  o.oSigT  liters.  This  term  ^^>  however, 


is  equal  to  (i+a)  (i.e.,  3.179:^:  :i  +  a  :  i)  ;    hence  V  = 
T(i+a)  and  V'  =  T'(i+a'),  and  we  obtain 


(i  -a'2)          ^  T(i  -a2) 
From  the  results 

r  =  273  +  26.i,        ^  =  2.65,      a 
^  =  273  +  111.3,     ^'  =  1.65,     a' 

we  find  q=  —  12900  cals. 


a2  q  /i       i\ 

i  -a2  =2~  (f~  T>  ' 


CHEMICAL  CHANGE.  261' 

After  subtracting  the  heat  equivalent  to  the  work  of 
expansion  *  we  obtain  from  the  experimental  result 

q=  —12500. 

By  a  similar  calculation  it  is  also  possible  to  find  the 
heat  of  dissociation  of  other  gases. 

B.  CHEMICAL  KINETICS 

63.  Application  of  the  law  of  mass  action.  —  Thus  far 
we  have  only  considered  the  equilibrium  which  is  attained 
after  the  reaction  has  come  to  rest,  i.e.,  after  the  two  sides 
bear  a  constant  relajtion  to  one  another.  The  question 
now  arises  as  to  the  progress  of  a  reaction  toward  this 
state,  and  the  factors  upon  which  the  time  necessary  to 
attain  it  depends. 

By  aid  of  the  law  of  mass  action,  it  is  possible  to 
find  an  answer  to  both  portions  of  this  question.  Since 
chemical  action  at  any  time,  according  to  it  (p.  224),  is 
proportional,  for  constant  temperature,  to  the  active 
masses  of  the  substances  present,  i.e.,  to  those  portions 
which  are  free  to  act,  then,  when  we  have  two  substances 
reacting,  the  concentrations  being  #1  and  a2  moles  per 
liter, 


*  The  gas  is  supposed  to  increase  its  volume  without  doing  work. 


262  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

where  x  is  the  fraction  of  a  mole  of  each  which  decom- 
poses in  the  time  /.  The  term  k  in  this  equation  is  known 
as  the  speed  constant  of  the  reaction,  and  is  constant  at 
any  one  temperature  for  any  value  of  x  in  the  reaction 
in  question. 

Suppose  we  have  the  reversible  reaction 


which  after  a  time  attains  a  state  of  equilibrium  in  which 
all  four  products  are  present.  The  relative  amounts  of 
these  are  dependent  upon  the  value  of  K  for  this  reaction 
at  this  temperature  according  to  the  relation 


If  we  start  with  a\  moles  of  A\  and  a2  moles  of  A^  then 
dx 


But  if  we  start  with  a\'  moles  of  A\  and  aj  of  AJ,  then 


dx' 
where  —rr  is  the  velocity  in  the  opposite  direction. 


CHEMICAL   CHANGE.  263 

Starting  with  the  substances  on  either  side,  then,  those 
on  the  other  will  exert  an  ever-increasing  influence  upon 
the  velocity  due  to  the  initial  substances,  and  this  velocity 
must  decrease  continually.  Finally,  however,  equilib- 
rium will  be  attained  and  the  ratio  of  the  amounts  on  the 
two  sides  will  remain  constant,  i.e.,  the  reaction  as  a 
whole  ceases,  and  any  motion  which  exists  is  so  compen- 
sated by  a  contrary  one  that  it  does  not  appear. 

Imagine  we  start  with  a^  moles  of  A\  and  a2  moles  of 
A  2.  The  total  velocity  due  to  these  at  any  one  time 
will  be 


7  y 

and  at  equilibrium,  i.e.,  where  -jr  =  o, 


or 


TT^  A  = 


i.e.,  the  equilibrium  constant,  K,  oj  any  reversible  reaction 
is  equal  to  the  ratio  oj  the  speed  constants  oj  that  reaction 
jor  the  two  directions. 

This  has  been  proven  to  be  true  for  a  number  of  cases. 
For  the  system  acid-alcohol  (p.  223)  it  was  found  for  a 
certain  strength  acid  that  £  =  0.000238  and  £'  =  0.000815, 


264  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

k 
from  which  K=-r^  =  2.g2,  while  direct   experiment   gave 

#  =  2.84. 

All  this  is  only  true,  however,  when  the  reaction  takes 
place  isothermally  ,  i.e.,  when  the  heat  liberated  or  absorbed 
is  removed  or  supplied  and  no  change  in  the  temperature 
results,  j  or  the  constants  k  and  k'  are  dependent,  upon  the 
temperature. 

In  general  the  application  of  this  formula  is  very  much 
simplified  by  the  fact  that  most  reactions  are  almost 
complete  in  one  direction,  so  that  the  second  term  will 
be  so  small  that  it  may  be  neglected.  We  have  then 

dx 


In  all  these  cases  the  values  on  the  right  are  obtained 
by  subtracting  the  loss  from  the  concentration  of  the 
original  substance  and  having  as  many  such  terms  as 
there  are  moles  in  the  formula.  Thus  for  the  reaction 
A  =  2B  +  D  we  would  write 

dx 

—  =k(cB-x)(cB-x)(cD-x). 

This  is  simply  custom  (see  p.  236),  for  we  could  also 
write  it 

-     =  k'(cB-2x)2(cD-x), 


CHEMICAL   CHANGE.  265 

and  although  the  k  value  would  be  different,  it  would 
be  constant. 

64.  Reactions  of  the  first  order.  —  For  convenience  we 
shall  divide  all  reactions  into  orders.  Thus  a  reaction 
of  the  first  order  is  one  in  which  but  one  substance  suffers 
a  change  in  concentration.  This  definition  is  to  be  fur- 
ther restricted,  in  that  for  the  first  order  it  is  necessary 
that  the  equation  show  but  i  mole  of  one  substance 
changing  its  concentration. 

Cane-sugar  in  water  solution  is  transformed  in  the 
presence  of  acids  almost  completely  into  dextrose  and 
laevulose;  it  is  inverted.  The  speed  of  the  reaction  is 
very  small  and  increases  with  the  amount  of  acid  added. 

The  progress  of  the  reaction  may  be  observed  by  aid 
of  the  polariscope.  The  unin  verted  portion  revolves 
the  plane  of  polarization  to  the  right,  while  the  two 
products  revolve  it  to  the  left.  If  «o  is  the  positive 
angle  of  revolution  at  the  time  /=o,  i.e.,  that  which 
is  due  to  the  uninverted  sugar  alone,  the  amount  of 
which  is  a  moles,  a0'  is  the  negative  angle  after  com- 
plete inversion,  and  a:  is  the  angle  at  the  time  /,  then, 
since  the  revolution  is  proportional  to  the  concentra- 
tions, x,  the  amount  inverted,  is  found  from 


x=a 


For   the   time   /=o°,  a=a^   i.e.,    #=o;     for   the   time 
/  -  oo  ,  i.e.,  after  complete  inversion,  a  =  —  an  or  x  =  a. 


266  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

This  process  was  first  measured  by  Wilhelmy  (1850), 
and  it  has  played  an  important  role  in  the  history  of 
chemical  mechanics. 

The  process  follows  the  scheme 

Ci2H22On  +H2O  =  2C6Hi2O6 

whether  acid  is  used  or  not,  for  the  concentration  of 
the  latter  does  not  change  during  the  reaction.  Accord- 
ing to  the  law  of  mass  action  the  speed  is  proportional 
to  the  amounts  of  sugar  and  water.  The  latter,  how- 
ever, is  present  in  such  an  excess  that  its  action  may 
be  regarded  as  constant.  The  speed  of  reaction,  then, 
is  proportional  to  the  amount  of  sugar  present  and  we 
have  a  reaction  of  the  first  order,  i.e.,  the  formula  is 

dx  . 


where,  for  t=o,  #=o,  and  k  is  the  inversion  constant, 
which  depends  only  upon  the  temperature.  By  integra- 
tion this  becomes 

—  loge  (a—  x)  =kt+  constant, 

or  since,  for  /=o,  #=o, 

(a)  =the  constant, 


i.e. 

a       i 


CHEMICAL   CHANGE.  267 

in  other  words,  a  constant  fraction  of  the  tota.  amount 
of  sugar  is  inverted  in  each  unit  of  time. 

The  meaning  of  the  constant  k  in  words  is  as  follows: 
Its  reciprocal  value  multiplied  by  the  natural  logarithm 
of  2  gives  the  time  in  minutes  which  is  necessary  for  the 
transformation  of  one  half  the  total  amount  of  substance, 
provided  the  products  of  the  reaction  are  removed  as  soon 
as  they  are  formed  and  the  substances  replaced  as  they  are 

used.  This  is  shown  by  the  substitution  of  — for  x.  Fur- 
ther, for  all  reactions  of  the  first  order,  this  constant  k 
is  independent  of  the  original  concentration  of  the  sub- 
stance. Table  VIII  gives  the  values  for  k  for  a  20% 
sugar  solution  in  presence  of  3,0.5  N.  solution  of  lactic 
acid  at  25°  C 


TABLE  VTTI. 

INVERSION   OF   SUGAR. 
/(minutes).  a. 


1435 
43i5 
7070 

"360 
14170 


19815 
29925 
«         -100.77 


OT      •  J 

3i°-i 

0.2348 

25°.0 

0-2359 

20°.  16 

0.2343 

13°.  98 

0.2310 

10°.  01 

0.2301 

7°-  57 

0.2316 

5°-o8 

0.2291 

i°.6S 

0.2330 

268  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Since  only  the  constancy  of  the  term  -  log^ is  to  be 

shown,  the  more  convenient  system  of  logarithms,  i.e., 
that  of  Briggs,  has  been  employed  above. 

65.  Catalytic  action  of  hydrogen  ions.  Catalysis.— 
The  inversion  of  sugar,  as  well  as  all  other  reactions  of 
the  first  order  which  are  hastened  by  the  action  of  acids, 
is  apparently  due  to  the  ionized  hydrogen  which  is  pres- 
ent. This  action  is  known  as  a  catalytic  action  of  the 
acid  present,  which  retains  throughout  the  reaction  its 
original  concentration.  The  reason  why  H*  ions  should 
act  in  this  way  is  unknown;  but  that  they  do  have  an 
accelerating  effect  is  an  established  fact. 

In  general  a  catalytic  action  is  one  that  hastens  the 
reaction  in  which  it  is  without  at  the  same  time  having 
its  concentration  changed  by  the  reaction.  An  example 
of  such  an  action  is  given  by  the  increased  speed  of  solu- 
tion of  metallic  mercury,  silver,  or  copper  in  a  nitric-acid 
solution  in  which  one  of  these  has  already  been  dissolved 
to  a  slight  extent.  The  catalytic  action  here  is  due  to 
the  formation  of  the  oxides  of  nitrogen,  these,  when 
passed  into  the  nitric  acid  from  an  exterior  source,  caus- 
ing the  same  action  as  the  above. 

Since  the  catalysor  takes  no  part  in  the  reaction  it  does 
not  alter  the  equilibrium  constant,  hence  it  must  also  has- 

k 
ten  the  reverse  action,  so  that  -77  may  remain  constant. 

»          K 


CHEMICAL   CHANGE.  269 

As  there  is  no  general  law  known  for  catalytic  action  the 
reader  must  be  referred  elsewhere  for  the  vast  number  of 
isolated  facts. 

The  more  concentrated  the  acid  used  the  more  rapidly 
the  sugar  is  inverted,  without,  however,  any  exact  propor- 
tionality. The  inverting  action  increases  more  rapidly 
than  the  concentration.  In  presence  of  a  neutral  salt  of 
the  acid  its  inverting  power  is  increased  by  about  10% 
for  the  stronger  acids,  but  decreased  for  the  weaker  ones. 

Since  the  acceleration  of  the  speed  of  the  reaction  is 
caused  only  by  acids,  and  since  they  are  distinguished  by 
the  presence  of  H*  ions,  the  action  must  be  due  to  these 
latter.  Further  than  this,  a  series  of  acids  arranged  in 
the  order  of  inverting  power  is  found  to  be  in  the  same 
order  in  which  they  stand  in  reference  to  ionization.  Thus 
it  is  plain  that  the  H*  ions  have  this  effect,  although  in 
concentrated  solutions  the  inverting  power  is  not  strictly 
proportional  to  the  ionization.  For  example,  a  0.5  N. 
solution  of  HC1  inverts  6.07  times  as  rapidly  as  one  of 
a  o.i  N.  solution,  while  it  contains  only  4.64  times  as 
much  ionic  H*.  Arrhenius  found  the  presence  of  other 
ions  to  increase  the  catalytic  action  of  those  of  H*. 

That  the  speed  of  reaction  increases  more  rapidly  than 
the  concentration  of  H'  ions  is  assumed  to  be  due  to  the 
fact  that  the  negative  ions  are  also  present  to  an  equal 
amount  and  so  increase  the  inverting  action  of  the  H'  ions. 

In  the  presence  of  a  neutral  salt,  then,  we  have  two 


270  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

separate  actions  upon  the  acid:  the  first  causes  a  de- 
crease in  the  concentration  of  H'  ions  of  the  acid  (see 
p.  144)  ;  the  second  accelerates  the  action  of  the  H'  ions 
left. 

The  stronger  acids  are  less  influenced  by  the  former 
than  the  weaker  ones,  because  they  are  dissociated  more 
nearly  to  the  extent  to  which  their  salts  are.  Consequently 
the  second  action  predominates,  and  we  have  the  greatest 
inverting  power  for  the  strong  acids  when  a  salt  is  present. 

For  the  weaker  acids,  however,  the  second  action  is  the 
smaller;  hence  the  inverting  action  is  less  for  weak  acids 
in  presence  of  a  salt. 

Palmaer  has  found  that  in  dilute  solutions,  when  the 
influence  of  the  neutral-salt  action  is  avoided,  the  inver- 
sion constant  is  exactly  proportional  to  the  concentration 
of  hydrogen  ions. 

Another  reaction  of  the  first  order  which  depends  for 
its  speed  upon  the  catalytic  action  of  the  H'  ions  is  the 
formation  of  alcohol  and  acid  from  an  ester.  For  example, 

CH3COOC2H5+H20  =  CH3COOH  +  C2H5OH. 

The  equation  is  the  same  as  before,  i.e.,  follows  the 
law 


where  a  is  the  amount  of  ester  in  moles  at  the  time 
/  =  o,  and  x  is  the  amount  transformed  at  the  time  /, 


CHEMICAL   CHANGE.  271 

and  is  equal  also  to  the  amount  of  acid  or  alcohol 
formed  by  the  reaction  during  the  time  /.  The  progress 
of  this  reaction  is  observed  by  titrating  the  amount  of 
free  acetic  acid  present. 

Ostwald  measured  the  value  for  the  reaction  with 
methylacetate.  He  started  with  i  c.c.  of  methylacetate 
and  10  c.c.  HC1  (i  mole  to  the  liter),  diluted  to  15  c.c. 
To  neutralize  this  acid  13.33  c-c-  °f  a  Ba(OH)2  solution 
would  be  necessary.  The  amounts  of  acetic  acid  formed 
are  shown  by  titration  after  substracting  13.33;  these 
amounts,  expressed  in  c.c.  of  Ba(OH)2  solution,  for 
various  times  are  as  follows: 

o  14  34  IQQ  539  QO          Minutes 

0.92     2.14    8.82     13.09     14.11      c.c. 

The  amount  of  ester  (a)  is  represented,  then,  by  14.11 
a  —  x  being  this  value  minus  0.92,  2.14,  etc.  The  values 
thus  determined  for  k  are  0.00209,  0.00210,  0.00214,  and 
0.00212. 

Another  reaction  which  gives  a  constant  when  con- 
sidered as  a  reaction  of  the  first  order  is  the  one  which  is 
usually  written  / 

4AsHs = As4  +  6H2. 

Since  at  310°  this  fulfils  the  definition  of  a  reaction  of  the 
first  order,  it  should  be  written  not  as  above,  as  of  the 
fourth  order,  but  as  of  the  first,  i.e. 


57$         ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

Here  also  the  velocity  is  exactly  proportional  when  the 
action  of  the  neutral  salt  is  avoided.  If  the  amount  of 
neutral  salt  in  the  solution  is  so  large  that  it  can  be  con- 
sidered constant,  then  a  constant  value  will  be  obtained 
for  k,  although  its  absolute  value  will  be  different  from 
that  for  the  case  that  no  neutral  salt  is  present.  Walker 
used  this  method  to -determine  H'  ions  present  in  a  solution 
due  to  hydrolytic  dissociation,  which  we  shall  discuss  in 
detail  below. 

66.  Reactions  of  the  second  order. — Here  two  mole- 
cules suffer  a  change  in  concentration  during  the  reaction, 
i.e.,  the  constant  depends  upon  the  concentration  of  two 
substances.  We  have  then 

dx 

—  =  k(a-x)(b-x), 

or 

— —-  [log,  (b  —  x)  —  log,  (a  -  x)]  =  kt  +  constant. 

For  t  =  o,  x  =  o,  and  the  constant  is 


i.e., 

,          i       .       (a-x)b 

&       (*       K\t  ^°&e  /A 


CHEMICAL  CHANGE.  273 

,•-  •  •  •  ••  — ^ 

If  we  use  equivalent  amounts  of  the  two  substances,  then 
a  =  b,  and  we  have 


or 

i       x 


t  (a—x)a 

Here  k  is  inversely  proportional  to  the  original  concentra- 
tion. 

An  example  of  a  reaction  of  the  second  order  is 

CH3CO(3C2H5  +  NaOH  =  CH3COONa  +  C2H5OH. 

Table  IX  gives  the  value  of  this  constant  as  calculated 
for  different  lengths  of  time  at  the  temperature  of  10°  C. 

TABLE  IX. 

. ,    .      .     N  Amount  NaOH  fc 

/(minutes)..          in  c.c.  of  acid.  *• 


0 

61.95 

489 

5°  -59 

2.36 

1037 

42.40 

2.38 

2818 

29-35 

2-33 

00 

14.92 

The  question  as  to  how  the  speed  varies  with  the  nature 
of  the  base  used  has  been  investigated  quite  thoroughly. 
Thus  Reicher  found  for  strong  bases  (those  which  are 
much  dissociated)  approximately  equal  values  for  k;  for 
weak  ones,  however,  he  found  very  much  smaller  values. 


274  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Ostwald  observed  that  the  weak  bases  ammonia  and 
allylamine  fail  to  give  a  constant  value  for  k.  Thus 
for  ethyl  acetate  and  ammonia  he  found  the  values  given 
in  Table  X. 

TABLE   X. 

/  (minutes).  k. 

O  

60  1.64 

*   240  I.O4 

1470  0.484 

This  failure  to  give  a  constant  he  recognized,  however, 
as  being  due  to  the  effect  of  the  neutral  salt  formed  (am- 
monium acetate)  upon  the  weak  base.  When  a  large 
amount  of  this  is  present  a  constant  is  obtained,  as  is 
shown  by  Table  XI. 

TABLE   XI. 

t  (minutes).  £. 

O  

994  0.138 

6874  o.i 20 

15404  0.119 

This  shows  still  a  slight  effect  of  the  neutral  salt 
formed,  for  it  is  impossible  to  keep  it  perfectly  constant. 

Arrhenius  found  that  the  base  acts  simply  in  pro- 
portion to  the  amount  of  ionized  OH'  it  contains,  i.  e., 
we  have 

C2H302C2H5  +  OH' + Na-  =  C2H3O2' + Na'  +  C2H6OH. 


CHEMICAL   CHANGS.  275 

According  to  this,  all  bases  containing  the  same  amount 
of  OH'  ions  should  give  the  same  constant.  We  can 
rearrange  our  equation,  then,  to  the  form 


where  a  is  the  degree  of  dissociation  of  the  base.  Since 
the  strong  bases  are  but  slightly  influenced  by  the  addi- 
tion of  salts  with  an  ion  in  common,  the  constant  of 
KOH  can  be  taken,  with  very  little  error,  to  be  that 
due  to  a  certain  concentration  of  OH'  ions.  From  this 
it  is  then  possible  to  calculate  the  constant  for  NH4OH, 
with  or  without  an  ammonium  salt  present.  A  m/^o  solu- 
tion of  KOH  at  24°. 7  C.  gives  a  constant  equal  to  6.41. 
Such  a  solution  of  NH4OH  is  but  2.69%  dissociated, 
while  that  of  KOH  is  97.2%  dissociated.  The  constant, 
then,  for  NH4OH  in  the  absence  of  salt  is 


0.177. 
0.972       ' 


The  effect  of  the  neutral  salt  upon  the  NH4OH  can 
be  readily  calculated  from  the  dissociation  of  the  two; 
we  have,  then,  the  constant  for  NH4OH  in  presence  of 
a  neutral  salt, 


276  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

where  a  is  the  concentration  of  OH'  ions  in  the 
NH4OH  in  presence  of  the  amount  of  the  neutral  salts, 
the  concentration  of  which  is  designated  below  by  s. 
Table  XII  compares  the  values  thus  obtained  with  those 
actually  observed  by  experiment. 

TABLE  XII. 

SPEED   OF   REACTION   CAUSED   BY  NH4OH. 


S 

a 

k  (calculated) 

k  (observei 

o 

2.69% 

0.177 

0.156 

0.00125 

I.  21 

0.08 

0.062 

0.005 

0.71 

0.047 

0.039 

0.0175 

O.IlS 

0.0078 

0.0081 

0.025 

O.O82 

0.0054 

o  .  0062 

0.05 

0.042 

0.0028 

0.0033 

This  is  a  very  good  agreement,  considering  the  pos- 
sible experimental  errors,  so  that  we  may  conclude  that 
the  •velocity  oj  saponification  is  proportional  to  the  con- 
centration oj  OH'  ions  present,  and  independent  of  the 
radical  from  which  they  are  split  off. 

67.  Reactions  of  the  third  order.  —  If  the  3  mole- 
cules are  present  in  equimolecular  amounts,  we  have 


V 


or 


dx 

dt 


i  x(2a— x) 
k=— 


t  2a2(a—x)2' 


CHEMICAL  CHANGE. 


One  reaction  of  this  order  has  been  studied  by  Noyes 
(Zeit.  f.  phys.  Chem  ,  16,  546,  1895).     T*he  reaction  is 

n 

2FeCl3  +  SnCl2  =  2FeCl2  +  SnCl4> 


of 


The  results,  starting  with  a  concentration  of  .025  moles 
of  each,  SnCl2,  FeQ3,  SnCl4  and  FeCl2,  were  as  follows: 


Time 

X 

a—x 

k 

2-5 

0.00351 

0.02149 

113 

3 

0.00388 

O.O2II2 

107 

6 

0.00663 

0.01837 

114 

ii 

0.00946 

0.01554 

116 

IS 

o.  01106 

0.01394 

118 

18 

0.01187 

0.01313 

117 

30 

0.01440 

0.01060 

122 

60 

0.01716 

0.00784 

122 

Average  =  1 16 

For  such  a  reaction  oj  the  jd  order  k  is  inversely  pro- 
portional to  the  square  of  the  original  concentration;  and  k 
jor  a  reaction  oj  the  nth  order  would  be  inversely  propor- 
tional to  the  (n  —  i)  power  oj  the  original  concentration. 

Up  to  the  present  no  reactions  of  an  order  higher 
than  the  third  have  been  found,  so  that  we  need  not 
consider  them. 

In  a  second  order  reaction  the  effect  of  either  con- 
stituent upon  the  velocity  should  be  the  same;  in  a 


278  ELEMENTS   OF  PHYSICAL  CHEMISTRY. 

reaction  of  the  third  order  the  effect  should  be  different, 
and  for  the  above  Noyes  found,  indeed,  that  an  excess  of 
ferric  chloride  has  a  greater  effect  in  hastening  the  reac- 
tion than  has  stannous  chloride. 

68.  Incomplete  reactions.  —  Thus  far  we  have  con- 
sidered the  speed  of  reactions  which  are  complete,  i.e., 
those  which  only  cease  when  all  the  original  substance 
has  been  transformed.  In  the  case  where  this  is  not 
true  it  is  necessary  to  use  the  equation  in  its  original 
form.  Such  a  case  is  the  one  already  studied, 


O, 


which  goes  with  a  certain  speed  until  two  thirds  of  the 
acid  and  alcohol  are  decomposed.  When  the  amount 
of  ester  is  equal  to  #,  then  (p.  263) 


When  we  start  with  i  mole  of  each  acid  and  alcohol 
and  have  no  water  or  ester  present,  and  k  and  V  are  the 
speed  components  in  the  two  directions,  we  have  as 
above  (pp.  262-263) 


CHEMICAL   CHANGE.  279 

By  observing  the  change  for  any  time  we  find 


k 

From  these  two  values  77  and  k  —  tf  we  can  find  k.      The 
K 

reaction  so  measured,  however,  does  not  give  a  con- 
stant value  for  k.  This  was  accounted  for  by  Knob- 
lauch. When  alcohol  and  acetic  acid  form  ester  and 
water  the  reaction  goes  faster  than  it  should,  accord- 
ing to  the  theory.  This  is  due  to  the  catalytic  action 
of  the  H'  ions  present  when  the  reaction  goes  in  the" 
one  direction.  If,  then,  the  concentration  of  H'  ions 
is  retained  the  same  throughout  the  reaction,  k  should 
be  constant,  which  has  been  found  by  experiment  to  be 
the  case. 

This  shows  the  importance  of  secondary  reactions, 
which  may  give  entirely  false  results  unless  accounted 
for. 

69.  Reactions  between  solids  and  liquids. — The  solu- 
tion of  a  substance  in  an  acid  depends  for  its  speed 
upon  the  surface  of  contact  between  the  two  and  upon 
the  strength  of  the  acid;  but  many  secondary  reactions 
can  take  place  here,  so  that  our  results  are  only 
approximate. 

If  we  assume  the  surface  to  be  so  large  that  it  changes 


280  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

but  slightly  during  the  reaction  we  may  consider  it  as 
constant;    we  have  then,  if  5  is  the  surface, 


where  x  is   the  amount  dissolved   in   the   time  /.     By 
integration  we  find 


t 


The  formula  in  this  form  was  found  by  Boguski  to  give 
constant  results,  within  the  experimental  error,  for  the 
solution  of  marble  in  acid.  Noyes  and  Whitney  have 
proved  that  the  speed  of  solubility  of  a  solid  substance 
in  any  instant  is  proportional  to  the  difference  between 
the  concentration  of  saturation  and  the  one  at  the  time 
of  the  experiment. 

70.  Speed  of  reaction  and  temperature.  —  Empirically 
it  has  been  found  that  the  speed  of  a  reaction  always  in- 
creases with  an  increase  in  the  temperature.  The  increase 
in  many  cases  is  very  great.  Thus  a  rise  in  tempera- 
ture of  30°  causes  the  speed  of  sugar  inversion  to  be 
five  times  as  great  as  before.  An  empirical  formula  was 
found  by  van't  Hoff  and  tested  by  Arrhenius.  We  have 


where  CQ  is  the  speed  constant  at  the  temperature  TQ, 

k.  •• 


CHEMICAL   CHANGE.  281 

Ci  is  the  constant  at  TI,  e  is  the  base  of  the  natural 
logarithms,  and  A  is  a  numerical  constant.  Using  the* 
lower  of  two  temperatures  as  T0,  we  have  for  the  trans- 
formation of  ammonium  cyanate  into  urea 

7^0  =  273  +  25°,     €"0  =  0.000227,     A  =  11700. 


Tl 

C(obs.) 

C  (calc.) 

273+39 

0.00141 

0.00133 

273+50.1 

0.00520 

0.00480 

273+64.5 

0.0228 

0.0227 

273+74.7 

0.062 

0.0623 

273+80 

O.IOO 

0.105 

C.  APPLICATION  OF  THE  LAW  OF  MASS  ACTION  TO 
ELECTROLYTES — IONIC  EQUILIBRIA. 

71.  Organic  acids  and  bases.  The  Ostwald  dilution 
law. — The  application  of  the  law  of  mass  action  to  gaseous 
equilibria,  as  well  as  to  those  existing  in  solutions  of  non- 
electrolytes — in  short  to  chemical  equilibria — has  shown 
that  it  is  a  general  law  of  nature,  holding  between  very 
wide  limits.  The  question  which  naturally  arises  here, 
then,  is,  Can  the  law  of  mass  action  also  be  applied  to 
those  equilibria  existing  between  'the  ionized  and  un- 
ionized portions  of  a  substance  in  solution?  In  other 
words,  is  the  law  of  mass  action  the  principle  govern- 
ing the  amounts  of  these  portions  which  can  exist  together 
hi  equilibrium?  It  is  the  purpose  of  this  section  to 


282  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

answer  this  question  so  far  as  is  possible  from  our  present 
knowledge  of  the  facts. 

The  conductivity  ratios  —  ,  as  well  as  the  methods  for 

jHoo 

determining  the  average  molecular  weight,  where  these 
can  be  carried  out  with  sufficient  accuracy,  show  that  a 
water  solution  of  acetic  acid  is  ionized  according  to  the 
following  scheme: 


From  this  equation,  applying  the  law   of  mass-action, 
we  obtain 


or  (see  page  234) 


which  is  known  as  the  Ostwald  dilution  law,  for  it  gives 
the  relation  of  ionization  to  dilution. 

Substituting   the    ratio    —  for  a:  at  various  dilutions, 

/*00 

Ostwald  found  K  to  be  constant,  with  a  value  at  25°  of 
i.SXio^5.  From  the  value  of  K  at  any  temperature, 
then,  it  is  possible,  by  solving  for  a,  to  find  the  degree 


CHEMICAL   CHANGE.  283 

of  ionization  at  any  dilution  or  at  any  concentration  at 
that  temperature,  for  c  =    .     We  find 


> 
2 


where  ^  is    the    equilibrium,  dissociation,  or  ionization 
constant,  or  the  so-catted  coefficient  oj  affinity,  of  the  acid. 

Further  than  this,  when  K  is  known  for  any  tem- 
perature it  is  possible  to  find  a,  at  any  dilution,  in  the 
presence  of  an  acid  or  salt  with  an  ion  in  common 
(H*  or  CHsCOO')-  And,  just  as  in  the  case  of  gaseous 
dissociation,  we  always  find  a  smaller  dissociation  in  the 
presence  of  the  products  arising  from  an  exterior  source. 
Naturally  the  calculation  here  is  similar  to  the  other 
(p.  238).  For  example,  we  have 


(CH-  +  x  -  y 


where  x  represents  the  concentration  of  H*  ions  due  to 
other  substances,  CH-  and  CcHaCOO'  are  the  concentrations 
due  to  the  acid  in  the  absence  of  other  substances,  and 
y  is  the  concentration  of  each  (H*  and  CHaCOO')  lost,  i.e., 
uniting  to  form  un-ionized  CH3COOH,  as  the  result  of 
the  presence  of  x  moles  of  H'  ions.  The  concentrations 
of  H*  and  CH3COO',  now  arising  from  the  acid,  are 


284  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

equal,  then,  to  (CH-  —y)  =  (^cH3coo'->');  and  the  un-ion- 
ized  acid  concentration  is  (^cH3cooH+>')- 

Indeed  everything  which  we  found  above  (pp.  228-241) 
to  hold  true  for  gaseous  dissociation,  with  the  one  ex- 
ception mentioned  below,  also  holds  true  for  the  relation 
of  ionized  and  un-ionized  portions  of  a  substance,  of  the 
type  of  acetic  acid,  in  solution.  And,  as  a  rule,  the  re- 
sults are  simpler  to  calculate,  for  the  volume  change  of 
the  system  is  so  small  as  to  be  negligible. 

We  can  also  define  the  ionization  constant  in  terms 
similar  to  the  definition  of  the  gaseous  dissociation  constant 
(p.  235).  Thus  by  multiplying  K  for  acetic  acid  by  2  we 
obtain  0.000x^36  as  the  concentration  in  moles  per  liter  of 
a  solution  of  acetic  acid  which  would  be  50%  ionized, 
i.e.,  a  solution  of  i  mole  of  acid  in  27777.5  liters  of  water. 

All  organic  acids  when  treated  in  this  way  give  a  con- 

a2 

slant  value  jor  the  expression ^-,.     This  is  the  dif- 

(i—  a)V 

ference  between  gaseous  and  electrolytic  dissociation 
equilibria,  at  least  when  the  latter  is  for  an  organic  acid. 
The  acid,  whether  it  be  mono,  di,  or  polybasic,  always 
ionizes  as  a  monobasic  acid  up  to  the  dilution  at  which 
a  =  0.5.  This  means  that,  assuming  the  acid  to  ionize 
simply  into  the  two  products  H*  and  the  negative  radical 
(i.e.,  for  the  calculation  of  /*oo  )>  which  may  also  contain 
replaceable  hydrogen,  a  constant  is  obtained  so  long  as 
the  ionization  in  this  way  is  50%  or  less.  Beyond  that 


CHEMICAL   CHANGE.  285 

point  the  H*  ions,  due  to  a  breaking  down  of  the  negative 
ion  containing  replaceable  hydrogen,  are  great  enough 
in  concentration  to  influence  the  constant,  which  begins 
to  vary.  Above  the  dilution  at  which  a  =  0.5  the  second 
and  following  replaceable  hydrogens  begin  to  appear  as 
ions,  and  must  be  taken  into  account;  and  at  infinite  dilu- 
tion, if  it  were  possible  to  attain  it,  the  polybasic  acid 
would  be  composed  of  all  the  replaceable  hydrogens  as 
ions,  together  with  the  negative  radical,  without  replacea- 
ble hydrogen,  as  the  negative  ion. 

The  calculation  of  the  degree  of  ionization  of  dibasic 
acids  above  the  dilution  at  which  a  =  0.5  has  been  worked 
out  by  W.  A.  Smith  (Zeit.  f.  phys.  Chem.,  25,  144,  and  193, 
1898).  For  a  tribasic  acid,  for  example,  we  have  the 
following  processes  taking  place: 


HA"  =  H'-{-A'", 

for  which,  by  application  of  the  law  of  mass  action,  we 
obtain 


a? 


a,,2 


286  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Up  to  the  dilution  at  which  01  =  0.5  the  constants  k2 
and  £3  are  so  small  that  in  general  they  have  no  effect 
on  the  concentration  of  H*  ions,  which  is  practically  equal 
to  CH-'-  Above  that  point  the  value  of  H*  at  any  dilution 
must  be  taken  as  equal  to  CH-'  +  CH>"+CK-'". 

The  values  of  the  constants  k<2  and  £3,  and  the  con- 
centrations CH-"  and  CH-'"»  can  be  obtained  by  observ- 
ing the  catalytic  action  of  the  acid  salts,  for  example 
NaH2A  and  Na2HA,  upon  sugar  at  100°  or  methyl  acetate 
(p.  265).  In  these  cases,  also,  we  should  have  the  relations 
given  above  for  k2  and  £3,  and  experiment  has  shown 
this  to  be  the  case.  Since  the  mono  and  disodium  salts 
of  a  tribasic  acid  increase  in  ionization  approximately 
proportional  to  the  volume,  the  concentration  of  H* 
in  acid  salts  is  almost  independent  of  the  dilution.  In 


, 
other  words,  since  we  have  k2  =  —  —  ,  and    CH.A' 

^H2A' 

is  approximately  constant   in  moles,  per  liter,  c"H-  must 
also  remain  constant. 

For  citric  acid  (tribasic)  the  following  values  at  25° 
were  determined  by  Smith  (I.e.)  : 


k'2=     3-2X10-5, 

&3=   0.07  Xio~5. 

From  these  values  the  actual  concentration  of  H' 
ions  can  be  calculated  at  any  dilution;  and  it  will  be 
found  that  the  second  hydrogen,  indeed,  has  little  effect 


CHEMICAL   CHANGE. 


287 


until  the  amount  of  H2A'  is  greater  than  one-half  the 
total  concentration.  For  further  details  respecting  di- 
basic acids  see  Wegscheider  (Sitzungsber.  d.  Akad.  d. 
Wissenschaft.  in,  441-510,  1902). 

A  few  ionization  constants  are  given  below  in  Table 
XIII;  they  all  refer  to  the  first  hydrogen,  i.e.,  are  equal 
to  K  in  the  above  s.t  of  formulas. 

TABLE  XIII. 

IONIZATION    CONSTANTS   OF   ORGANIC    ACIDS    AT    25°  C. 

KX  10* 

Malic 39.5 

Fumaric 93 

Tartaric 97 

Salicylic 102 

Orthophthalic 121 

Monochloracetic 155 

Malonic 158 

Maleic 1170 

Dichloracetic 5140 

Oxalic 10060  (±)* 

Trichloracetic 121000  (±)* 


•34 
•44 
•45 
•49 
.61 
.80 


KX 

Propionic 

Isobutyric 

Capronic 

Butyric 

Valerianic 

Acetic 

Camphoric 2.25 

Anisic 3 .20 

Phenylacrylic 3.55 

Succinic 6. 65 

Lactic 13.8 

Glycollic 15.2 

Formic 21.4 


The  value  of  K  at  18°  for  a  few  other  acids,  some  of 
which  are  inorganic,  but  characterized  by  their  small 
ionization,  and  by  obeying  the  law  of  mass  action,  are 
given  by  Walker  and  Cormack  (Trans.  Chem.  Soc., 
77,  8,  1900),  the  value  of  acetic  acid  being  given  for 
comparison. 

*  These  two  acids  are  so  largely  dissociated  that  a  small  error  in  a 
affects  K  to  a  large  degree.  For  a  list  containing  a  very  large  number 
of  other  acids  see  Zeit.  f.  phys.  Chem.,  3,  418-20,  1889,  or  Kohl- 
rausch  and  Holbora,  Leitvermogen  der  Elektrolyte,  pp.  176-193. 


288  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

Per  Cent 


Name.  XXio» 

Solution. 

Acetic  acid  (25°),  CH3COO'  —  H'  .............  1,800,000  i  .  3 

Carbonic  acid,  H'  —  HCO3'  ..................         3,040  o  .  174 

H-—  C03"  ....................  0.6*  ..... 

Hydrogen  sulphide,  H'  —  HS'  ..................  570  0.075 

Boric  acid,  H'  —  H2BO3'  .....................  17  0.013 

Hydrocyanic  acid,  H'  —  CN'  ..................  13  o  .01  1 

Phenol,  H-—  C2H6C"  ......  .  .................  1.3  0.0037 


As  the  result  of  his  work  on  organic  acids  Ostwald 
formulated,  among  others,  the  following  law,  which  will 
serve  as  a  means  of  foreseeing  the  value  of  the  constant 
of  a  substituted  acid  when  that  of  the  simple  acid  is  known. 

The  substitution  of  O,  Cl,  Br,  I,  CN,  etc.— in  short, 
of  negative  groups — in  a  molecule  increases  the  separa- 
tion of  ions  of  H  in  water  solution;  the  addition  oj 
H,  NH2,  etc.,  i.e.,  positive  groups,  decreases  it;  and  the 
influence  is  the  greater  the  nearer  (in  space)  the  substi- 
tuted group  is  to  that  of  carboxyl.  Thus 

for  CH3COOH      100^  =  0.00180, 
forCC!3COOH      100^=121. 

In  the  case  of  the  addition  of  a  salt  with  an  ion  in 
common  to  an  organic  acid  (p.  283)  the  following  will 
be  seen  at  once  to  be  true.  If  the  degree  of  dissociation 

*  McCoy,  Am.  Chem.  Jour.,  29,  455,  1903. 


CHEMICAL   CHANGE.  289 

of  a  salt,  with  an  ion  in  common  with  an  acid,  is  d,  and  n 
is  the  number  of  moles  of  salt  which  are  present,  then  the 
equation  of  equilibrium  of  the  acid  will  become 


For  very  weak  acids  we  can  generalize  this  as  follows: 
a,  the  degree  of  dissociation  of  the  acid,  is  very  small 
in  presence  of  the  salt,  so  that  a  in  comparison  to  i  and 
nd  may  be  neglected.  Since  d  for  salts  is  almost  inde- 
pendent of  the  dilution,  we  have 


KV 
- 

n 


i.e.,  the  dissociation  of  a  weak  acid  in  presence  of  one 
of  its  salts  is  approximately  inversely  proportional  to 
the  amount  of  the  salt. 

The  case  of  the  partition  of  a  base  between  two  acids 
depends  to  a  certain  extent  upon  their  ionization  con- 
stants. This  partition  takes  place  when  there  is  not 
enough  base  present  to  saturate  both  acids.  The  final 
mixture  consists  of  water,  undissociated  salt,  and  the 
dissociated  and  undissociated  portions  of  the  acids. 
The  equilibrium  is  the  same  as  that  which  would  be 
attained  by  the  mixture  of  the  salt  of  the  one  acid  with 
the  other  acid.  The  affinity  of  each  acid  for  the  base 


290  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

will  depend  upon  the  percentage  of  free  ions  of  H*  which 
it  possesses,  i.e.,  the  one  containing  the  larger  quantity 
will  unite  with  the  larger  amount  of  the  base. 

For  weak  acids  it  is  possible  to  formulate  a  general 
law  regarding  the  partition  and  the  ionization  constants. 
In  this  case  a  is  so  small  that  it  may  be  neglected  in 
i  —  a  ;  hence  we  have 


a  =  VKV. 
And  for  two  acids  at  the  same  dilution 


or 

a 

1y 


The  coefficient  oj  partition  oj  two  acids,  then,  is  propor- 
tionate to  the  ratio  0}  their  degrees  of  dissociation  at  the  given 

VK 

volume,  or  jor  WEAK  acids  to  -7=. 

v  Kr 

This  coefficient  of  partition  is  independent  of  the 
nature  of  the  base  and  depends  only  upon  the  two  acids. 

For  the  partition  of  an  acid  between  two  bases  the  co- 
efficient will  depend  only  upon  the  two  bases.  If  a  is 


CHEMICAL   CHANGE.  291 

the  degree  of  dissociation  of  the  one  base  and  a'  that  of  the 
other,  we  shall  have  for  them  when  weak,  just  as  for  acids, 


a 


VJt 


and  the  same  generalization  holds  true. 

In  order  that  a  base  may  be  divided  equally  between 
two  acids  it  is  necessary  that  they  be  isohydric,  i.e.,  it  is 
necessary  that 


An  example  of  isohydric  solutions,  i.e.,  two  which 
contain  the  same  concentration  of  H*  ions,  is  acetic  acid 
at  a  dilution  of  8  liters  and  hydrochloric  acid  at  one  of 
667  liters.  These  two  solutions  may  be  mixed  in  all 
proportions  without  any  change  in  the  dissociation 
resulting.  When  mixed  in  equal  volumes,  if  treated 
with  a  small  amount  of  base,  equal  amounts  of  chloride 
and  acetate  will  be  formed. 

All  these  conclusions,  except  that  with  regard  to 
substitution  (p.  288),  hold  also  for  the  organic  bases,  as 
well  as  some  of  the  inorganic  ones.  The  ionization 
constants  for  these  are  found  from  the  relation 


~ 

A  = 


292  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

where  M*  is    the  positive  and    OH'  the  negative  ion. 
The  value  of  K  for  a  few  bases  is  given  below: 

Urea,  (4Q°.2) ,.  .  .K  =  0.0037X10-" 

Acetamide,      (4o°.2) o.oo33Xio~u 

Acetanilide,     (4o°.2) o  .OO44X  io~u 

Aniline,            (25°) 5.4       X  lo"10 

/>-Toluidine,    (4o°.2) 20.7       Xio"10 

w-Nitraniline,(4o°.2) 4           X  io~12 

p-         "           (4o°.2) i            Xio-12 

o-         "            (4Q°.2) o.oi      Xio-12 

Ammonium  hydrate,  NH;  — OH'  (25°) 23           X  10 

Methylamine,          (25°) 5           X 10 

Dimethalamine,      (25°) o  .074  X  io~ 

Trimethylamine,    (25°) o.oo74X  10 

Propylamine,           (25°) o  .047    X  io~ 

Isopropylamine,      (25°) °-°S3   X  io~ 

Isobutylamine,        (25°) 0.034  X  io~ 

The  equation  on  page  283  has  also  been  used  to  de- 
termine the  concentration  of  ions  in  the  solution  added  to 
a  substance  obeying  the  law  of  mass  action,  and  with 
very  good  results.  Thus,  starting  with  acetic  acid,  know- 
ing the  concentration  of  H"  ions,  and  adding  sodium 
acetate,  we  can  find  the  concentration  of  the  CH3COO' 
ions  in  the  salt.  In  this  case  the  new  concentration  of 
H*  ions  can  be  determined  from  the  speed  of  inversion 
of  sugar;  and,  by  solving  the  equation  for  the  amount 
of  CH3COO'  added,  we  can  find  the  dissociation  of  the 
salt  solution  added. 

This  same  process  may  also  be  carried  out  with  other 
systems,  mentioned  below,  and  by  it  it  is  possible  to 
determine  the  concentration  of  any  one  kind  of  ions.  As 


CHEMICAL   CHANGE  293 

mentioned  above,  the  ionization  has  been  found  to  be  sen- 
sibly the  same,  by  whatever  method  it  may  be  determined. 
72.  Acids,  bases,  and  salts  which  are  ionized  to  a  con- 
siderable extent.  Empirical  dilution  laws. — The  applica- 
tion of  the  law  of  mass  action  (the  Ostwald  dilution 
law)  to  the  above  so-called  strong  electrolytes  does  not 
lead  to  a  constant  value  for  K  when  the  ionization  is 

taken  from   the  conductivity  ratio   - — .     As  in  general 

;j.*> 

the  degree  of  dissociation  is  jound  to  be  the  same  by  all  of 
the  possible  methods,  our  only  conclusion  at  present  is 
tliat  the  law  oj  mass  action  cannot  be  applied  to  the  equi- 
librium oj  un-ionized  and  ionized  portions  in  such  solutions 
as  these.  Although  this  is  true  with  regard  to  very 
soluble  salts,  there  is  a  quantitative  relation,  which  we 
shall  develop  below,  holding  for  the  ionized  portions  of 
difficultly  soluble  substances,  when  the  un-ionized  por- 
tion is  retained  constant. 

The  only  known  case  of  a  dissociating  and  very 
soluble  salt  to  which  the  law  of  mass  action  may  be 
applied,  i.e.,  the  only  exception  to  the  above  conclusion, 
is  caesium  nitrate,  when  a  is  determined  by  the  freezing- 
point  method.  And  this  is  not  true  for  a  as  determined 
by  the  other  methods.  In  other  words,  in  this  case  the 
freezing-point  results  point  to  a  different  degree  of  ion- 
ization than  any  other  method.  The  results  for  caesium 
nitrate  as  determined  by  Biltz  (Zeit.  f.  phys.  Chem.,  40, 


294 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


218,  1902)  by  the  freezing-point  and  conductivity  meth- 
ods are  given  below. 

BY  DEPRESSION  or  THE  FREEZING-POINT. 


c  (Moles 
per  Liter). 

V  (Liters 
per  Mole). 

Depres- 
sion. 

a(0bs.). 

if        °2 

K  =-.  r- 

(i-a) 

0.00766 

l3°-7 

0.028 

0.98 

0-33 

0.01940 

51.6 

0.070 

0-95 

o-35 

0.04648 

21.6 

o.  164 

0.907 

[0.41] 

0.09884 

IO.2 

0-331 

O.SlO 

o-34 

o.  1421 

7.09 

0.460 

0.750 

0.32 

O  .  2  100 

4.8l 

0.662 

0.704 

0-35. 

0.2987 

3.385 

0.907 

0.641 

0-34 

0.3861 

2.62 

1.125 

0-575 

[0-30] 

0-4339 

2-33 

1.267 

0.578 

0-34 

Average  K,  excluding  values  in  brackets,  =  o .  34 


BY  CONDUCTIVITY,  /*oo  =  i47- 


V  (Liters  per  Mole). 

1024 

512 

256 

128 

64 

32 

16 

8 

4 


Vv. 

146.4 
144.7 
141.9 
139-3 
I38-3 
134.2 
I28.I 
I2O-9 
III.9 


0.14 

0.23 
0.30 
0-37 
0.47 
0.61 


Biltz  attributes  the  failure  of  the  law  of  mass  action, 
as.  applied  to  strong  electrolytes,  to  a  hydration  of  the 
substance — i.e.,  to  a  chemical  reaction  between  the 
substance  and  the  solvent — which  would  remove  active 
solvent  from  the  solution,  so  that  it  would  be  really  more 


CHEMICAL   CHANGE.  295 

concentrated  than  it  appears  to  be.  Just  why  the  solution 
of  caesium  nitrate  in  water  should  give  a  constant  value 
for  K  when  a  is  determined  by  the  freezing-point  method, 
while  for  conductivity  it  fails  to  do  so,  is  unknown  and 
thus  far  nothing  but  assumption  has  been  possible.  It 
is  to  be  remembered,  however,  that  this  value  of  K, 
although  giving,  when  solved  for  a,  the  ionization  accord- 
ing to  the  freezing-point  method,  does  not  give  the  value 
as  determined  by  any  other  method,  so  that  too  much 
stress  is  not  to  be  laid  upon  it,  especially  in  face  of  the 
fact  that  the  law  cannot  be  applied  to  any  other  strong 
electrolyte  by  any  method  of  determining  a.  It  would 
seem  more  probable  that  some  secondary  action  takes 
place  in  the  freezing  which  is  absent  in  all  other  cases, 
so  that  in  this  one  case  the  freezing-point  result  is  in- 
correct. This,  of  course,  may  not  be  true,  but  at  the 
present  time,  in  the  absence  of  any  indication  of  such  a 
result  for  other  substances,  it  is  decidedly  the  most 
reasonable  and  logical  one.  In  other  words,  then,  so 
far  as  we  know  at  present,  in  all  cases  but  this  the  con- 
ductivity leads  to  the  true  value  for  a,  so  that  it  is  but 
reasonable,  until  further  evidence  is  at  hand  to  assume 
it  be  correct  here,  and  attribute  the  case  above  to  some 
abnormality  not  yet  encountered  with  other  substances. 
Although  the  Ostwald  dilution  law  fails  utterly  to  hold 
for  the  equilibrium  between  the  ionized  and  un-ionized 
portions  of  strong  electrolytes,  certain  other  empirical 


296  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

dilution  laws  have  been  found  which  allow  us  to  find 
the  respective  amounts  at  any  dilution.  Thus  Rudolphi 
(Zeit.  f.  phys.  Chem.,  17,  385,  1895)  found  a  dilution 
law  which  gives  a  constant  value  between  certain  limits 
for  such  solutions  as  do  not  follow  the  Ostwald  dilution 
law.  This  law  is 

a2 

7 777=  =  constant, 

(i-a)v  V 

where  the  value  of  the  constant  is  approximately  the 
same  for  analogous  substances,  van't  Hoff  (Zeit.  f. 
phys.  Chem.,  18,  300,  1895)  altered  this  to  the  form 


a* 

-f ^r,  =  constant, 

(i  —  a)2V 

which   holds   even   better   than   Rudolphi's.     Simplified 
this  relation  is 

c? 

—^  =  constant, 


i.e.,  the  cube  o)  the  concentration  of  ions  divided  by  the 
square  oj  the  undissociated  portion  is  a  constant. 

Writing  the   Ostwald   dilution   law  in   this   form  we 
obtain  (for  binary  electrolytes) 


CHEMICAL   CHANGE.  297 

while  this  empirical  relation  (in  either  form)  remains  the 
same  jor  binary  or  ternary  substances;  in  other  words,  is 
independent  0}  the  number  oj  ions  formed  jrom  one  mole 
of  the  substance. 

Some  values  found  by  use  of  vant  Hoff's  equation 
are  given  below.  The  value  of  k  by  the  Ostwald  dilu- 
tion law  will  be  found  to  vary  considerably,  while  k^  is 
quite  constant,  i.e.,  within  the  experimental  error. 

AgN03  25°. 

F=  16  0=0.8283  £H  =  I.H 

=  32  =0.8748  =1.16 

=  64  =0.8993  =1.06 

=  128  =0.9262  =1.07 

=  256  =0.9467  =1.08 

=  512  =0.9619  =1.09 

Bancroft  (Zeit.  f.  phys.  Chem.,  31,  188,  1899)  pro- 
poses a  dilution  law  of  the  form 


constant  = — , 

in  which  the  constant  and  n  are  functions  of  the  nature 
of  the  electrolyte.  Although  this  relationship  is  as  yet 
unknown,  Bancroft  suggests  that  we  may  sometime 
find  a  relation  of  the  form 

w=2-/  (constant) 


298  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

(where  /  (constant)  varies  between  zero  and  a  value 
approximating  one-half)  which  will  reconcile  the  Ost- 
wald  dilution  law,  holding  only  for  organic  acids  and 
bases,  and  in  which  n=2,  with  this  form  in  which  n 
varies  between  the  values  1.36  and  1.5.  For  KC1  at 

c  -1'36 
18°  we  have  the  relation  -    —  =  2.63,  which  holds  in  a 

Cs 

very  remarkable  way  between  the  volumes  0.3  and 
10,000  liters,  as  will  be  seen  by  referring  to  the  results  in 
the  following  table: 


KC1  AT  1  8°. 

c.l.*9 

a 

Obs. 

Calc. 

0.987 

0.987 

0.984 

0.983 

0.978 

0.976 

0-973 

0.970 

0.965 

0.962 

0.950 

0.948 

0-934 

0-934 

0.915 

0.917 

0.902 

0.906 

0.883 

0.892 

0-853 

0.864 

0.821 

0.834 

0.803 

0-813 

0.780 

0.786 

0.748 

0-745 

0.706 

0.700 

0.673 

0.672 

IOOOO 

5000 

2OOO 
IOOO 

500 

200 
IOO 

SO 

33-3 

20 
10 

5 
3-3 

2 

I 
0-5 


C  >n 


C  > 

For  other  substances  the  relation   ~  is  not  so  satis 

c* 


CHEMICAL   CHANGE.  299 

factory  or  constant.  Noyes  (J.  Am.  Chem.  Soc.,  26, 
1 68,  1904)  has  determined  the  degree  of  ionization, 

— ,  for  the  chlorides  of  potassium  and  sodium  at  various 

r-<x> 

temperatures  and  finds  a  constant  value  for  the  ratio 
— j— ;  that  is,  the  fraction  of  salt  un-ionized  is  directly 

proportional  to  the  cube  root  of  the  concentration,  or  the 
concentration  of  un-ionized  substance,  (i—a)c,  is  directly 
proportional  to  the  4/3  power  of  the  total  concentration,  c, 
oj  salt.  The  degrees  of  ionization  of  potassium  and 
sodium  chlorides  were  found  to  be  nearly  identical  (the 
extreme  variation  being  2%)  at  all  temperatures  and 
dilutions.  In  a  o.i  molar  solution  the  dissociation  has 
approximately  the  following  values: 

18°  84%  281°  67% 

140°  79%  3o6°  60% 

218°  74% 


The  values  of  K'  =  — ^—  are  as  follows: 


18°  140°  218°  281°  306° 

NaCl 0.366          0.448          0.573          0.745          0.877 

KC1 0.321          0.468          0.577          0.713          0.853 

Some  of  the  observed  facts  as  to  the  ionization  of 
strong  electrolytes,  as  summarized  by  Noyes  (Tech- 
nology Quarterly,  17,  307,  1904)  are  as  follows: 

The  form  of  the  concentration  function  is  independent 


300  ELEMENTS  OF  PHYSICAL   CHEMISTRY, 

of  the  number  of  ions  into  which  the  mole  of  salt  disso- 
ciates. Instead  of  being  proportional  for  di-ionic,  tri- 
ionic,  and  tetra-ionic  to  the  square,  cube,  or  fourth 
power  of  the  concentration  of  the  ions,  the  undissociated 
portion  is  approximately  proportional  to  the  3/2  power 
of  that  concentration,  whatever  may  be  the  type  of  salt. 

The  conductivity  and  freezing-point  depression  of  a 
mixture  of  salts  having  an  ion  in  common  are  those 
calculated  under  the  assumption  that  the  degree  of 
ionization  of  each  salt  is  that  which  it  would  have  if 
present  alone  at  such  an  equivalent  concentration  that 
the  concentration  of  either  of  its  ions  were  equal  to  the 
sum  of  the  equivalent  concentrations  of  all  the  positive 
or  negative  ions  present  in  the  mixture.  Suppose  that 
a  mixed  solution  is  o.i  molar  with  respect  to  sodium 
chloride  and  0.2  molar  with  respect  to  sodium  sulphate, 
and  that  it  is  0.18  molar  with  reference  to  the  positive 
or  negative  ions  of  these  salts.  The  principle  then 
requires  that  the  ionization  of  either  of  these  salts  in 
the  mixture  be  the  same  as  it  is  in  water  alone  when 
its  ionic  concentration  is  0.18  molar.  This  has  been 
proven  conclusively  for  many  mixtures. 

The  decrease  of  ionization  with  increasing  concentra- 
tion is  roughly  constant  in  the  case  of  different  salts  of 
the  same  type. 

The  un-ionized  fraction  of  any  definite  molal  concen- 
tration is  roughly  proportional .  to  the  product  of.  the 


CHEMICAL   CHANGE.  3O1 

valences  of  the  two  ions  in  the  case  of  salts  of  different 
types.* 

From  these  facts,  together,  with  others,  Noyes  (l.c.) 
concludes  that  the  form  of  union  represented  by  the 
un-ionized  portion  of  a  substance  differs  essentially  from 
ordinary  chemical  combination,  it  being  so  much  less 
intimate  that  the  ions  still  exhibit  their  characteristic 
properties,  in  so  far  as  these  are  not  dependent  upon 
their  existence  as  separate  aggregates.  In  other  words, 
the  law  of  mass  action  is  inapplicable  to  the  relation 
between  ionized  and  un-ionized  portions  as  they  exist 
in  strong  electrolytes,  and  hence  this  is  not  to  be  con- 
sidered as  a  simple  chemical  equilibrium,  for  which, 
as  we  know,  the  law  of  mass  action  appears  to  hold 
rigidly. 

I  /73.  Heat  of  ionization.  —  By  van't  Hoff's  equation  it 
ns  possible  to  calulate  the  heat  of  ionization  of  a  sub- 
stance, provided  we  know  the  degree  of  dissociation  at 
two  different  temperatures.  This  is  true  not  only  for 
those  substances  which  follow  the  Ostwald  dilution  law, 
but  for  all  others  as  well,  according  to  Arrhenius  (Zeit. 
f.  phys.  Chem.,  4,  96,  1889,  and  9,  339,  1892).  For 
binary  electrolytes  we  have  then 

K'  a2  (i  -a)       q/i       i 

- 


*  Other  generalizations  of  this  kind  will  be  found  in  Chapter  IX 
under  Electrical  Conductivity. 


302  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

where  the  values  of  K  are  for  the  same  dilution  and 
represent  the  change  of  ionization,  even  though  the  val- 
ues are  not  the  same  as  other  dilutions,  and  the  values 
V  cancel  and  need  not  be  considered,  q  is  then  the  heat 
liberated  when  a  mole  of  substance  is  formed  in  solu- 
tion by  the  union  of  its  ions.  It  will  be  observed  here 
that  these  values  differ  from  those  calculated  from  the 
table  given  on  page  220,  for  these  refer  to  the  heat  of 
formation  of  the  ions  from  substance  already  in  solu- 
tion, while  those  refer  to  the  compound  process  of  solu- 
tion and  ionization,  i.e.,  the  difference  in  energy  between 
the  ionized  state  and  the  solid  or  gaseous  state.  Some 
of  the  results  as  found  by  Arrhenius  are  given  below, 
the  unit  being  the  small  calorie. 

HEATS  OF  IONIZATION. 


{ 


Substance.                                  Temperature.  Calories. 

-3* 

Propionic  acid  .................  •!  35Q  ~   557 

Butyric  acid  ............  {  ^  =   935 

Phosphoric  acid..                        ..{3£s  -^5» 

Hydrofluoric  acid  ...............     33°  —  3549 

Potassium  chloride  .............     35°  —   362 

iodide  ...............       "  -   916 

"          bromide  .............      "  —   425 

Sodium  chloride  ................      "  —  454 

"       hydrate  ................      "  —1292 

'  '       acetate  .................      "  —  391 

Hydrochloric  acid  ..........  ....      "  -  1080 


CHEMICAL  CHANGE. 

HEAT  NECESSARY  TO  COMPLETE  THE  IONIZATION,  (i  —  a)w;  (i  mole  in 

200  moles  of  water). 

Substance.                                  Temperatflre.  Calories. 

Potassium  bromide  ..............    35°  —     58 

"          iodide  ................    "  -   132 

chloride  ..............    "  56 

Sodium  hydrate  .................    "  —   180 

"       chloride  ................  .'    "  -     81 

Hydrochloric  acid  ...............     "  —    136 

Hydrofluoric  acid  ................   33°  —  3304 

Phosphoric  acid  .................   2i°.5  —  1682 

For  the  temperatures  of   35°  in  the  table,  7^=273  +  18, 
for    2i°.,    r 


Kf 

the  \oge  -g-  formula. 

From  data  such  as  the  above  it  is  possible  to  calculate 
the  heat  of  neutralization  of  an  acid  by  a  base.  The 
formula  for  this  (p.  217)  is 


where  the  figures  i,  2,  and  3  refer  respectively  to  acid, 
base,  and  salt,  and  x  is  the  heat  of  formation  of  i  mole 
of  water  from  H*  and  OH'  ions,  i.e.,  13,700  cal.  In  the 
table  below  the  calculated  values  of  q  at  two  tempera- 
tures are  given,  together  with  the  observed  values  at 
one  of  the  temperatures. 

It  is  obvious  from  the  results  above  that  the  value  of 
the  heat  of  neutralization  of  an  acid  by  a  base  cannot  be 
considered  as  indicative  of  the  strength  of  the  acid.  The 
two  latter  are  relatively  weak  acids  and  yet  they  give 
rise  to  the  greatest  amount  of  heat. 


304  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

HEAT  OF  NEUTRALIZATION  OF  ACIDS  WITH  NaOH. 
(i  mole  of  acid  + 1  mole  of  NaOH  +  4oo  moles  of  H2O.) 
At  35°.  At  21.5. 

Calc.  Caic.  Obs. 

HC1 12867  13447  13740 

HBr 12945  13525  13750 

HNO3 :   12970  JSSS0  13680 

CH3COOH 13094  13263  13400 

C2H6COOH 13390  13598  13480 

CHC12COOH 14491  14930  14830 

HgPO, 14720  14959  14830 

HF* 16184  16320  16270 

The  calculation  by  van't  Hoff's  formula  (pp.  252-261) 
of  the  heat  of  solution,  as  has  already  been  observed, 
is  satisfactory  when  the  substance  can  be  considered  as 
either  completely  ionized  or  completely  un-ionized.  The 
next  question  to  be  considered  is  the  calculation  of  the 
heat  of  solution  of  those  substances  which  are  but  par- 
tially ionized.  Unfortunately  no  relation  has  been  found 
which  gives  results  agreeing  very  closely  with  those  of  ex- 
periment. There  are  two  formulas,  however,  which  may 
be  used  together,  for  the  experimental  result  is  usually 
found  to  lie  midway  between  them,  i.e.,  one  gives  values 
which  are  too  high,  the  other  values  which  are  too  low- 
van't  Hoff's  formula  (p.  253)  leads  in  such  a  case  to 


*  For  data   as  to  the  ionization   of.  HF,  see  Deussen,  Zeit.  f.  anorg. 
Chem.,  44,  408,  1905. 


CHEMICAL   CHANGE.  305 

for  binary  electrolytes.     The  forms  given  by  Van  Laar 
are  two,  one  based  on  the  Ostwald  dilution  law,  viz., 


the  other  on  the  Rudolphi  dilution  law  (p.  296), 


As  the  results  by  these  laws  are  not  in  accord  with  expe- 
rience, and  we  could  hardly  expect  otherwise  with  the 
present  slight  knowledge  of  the  equilibrium  between 
the  ionized  and  un-ionized  portions,  we  shall  not  con- 
sider them  further,  for  it  is  quite  evident  that  until  we 
find  a  dilution  law  which  will  hold  for  all  dilutions  we 
cannot  hope  to  follow  closely  any  relation  depending,  as 
this  does,  entirely  upon  the  degree  of  ionization. 

74.  Solubility  or  ionic  product.  —  Although,  as  we 
have  seen,  the  law  of  mass  action  cannot  in  general  be 
applied  to  the  equilibrium  of  the  ionized  and  un-ionized 
portions  of  a  substance  in  solution  (except  to  organic 
acid  and  bases),  it  can  be  applied  with  considerable 
accuracy  to  a  very  large  number  of  saturated  solutions. 
An  example  of  such  an  equilibrium  is  a  saturated  solution 
of  silver  chloride,  which  is  found  to  be  practically  com- 
pletely ionized  according  to  the  scheme 

AgCl-Ag'  +  Cl'. 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 
Applying  the  law  of  mass  action  to  this  we  obtain 

™2 

Kc  =  c\c^     or    K  = 


(i-a)F' 

when  c  is  the  concentration  of  un-ionized  AgCl,  c\  that  of 
Ag',  and  c2  that  of  Cl'  ions.  Since  the  solution  is  satu- 
rated, the  value  of  c  at  any  temperature  must  remain  con- 
stant, for  if  the  solution  were  unsaturated,  solid  would 
dissolve,  if  supersaturated,  solid  would  precipitate.  We 
have  then,  at  any  one  temperature,  in  a  saturated  solu- 
tion of  silver  chloride,  the  relation 

Kc  =  constant  =  c\c2\ 

i.e.,  in  a  saturated  solution  of  a  binary  electrolyte  (of  this 
kind)  the  product  of  the  concentrations  0}  the  ions  must 
remain  constant,  with  unchanged  temperature. 

Expressing    this   in   a  more   general   form,   we    have 
for  the  reaction 


in  a  saturated  solution, 

(46)  Cin'c2n2  ^  constant  =  s, 

where  5  was  called  by  Ostwald  the  solubility  product 
of  the  substance.  This  solubility  product  is  of  para- 
mount importance  in  analytical  chemistry,  jor  a  precipi- 


CHEMICAL   CHANGE.  3°  7 

fate  (when  due  to  an  ionic  reaction,  and  most  oj  them 
can  be  shown  to  be  due  to  this)  is  always  and  only  formed 
when  its  solubility  product  is  exceeded.  This,  of  course, 
presupposes  that  no  supersaturation  phenomenon  is 
possible;  if  it  is,  then  the  metastable  limit  (p.  128)  must 
first  be  exceeded. 

Just  as  we  found  a  decrease  in  the  dissociation  of 
a  gas  or  an  organic  acid,  by  the  addition  of  one  of  the 
products  of  dissociation  from  an  exterior  source,  so  here 
the  addition  of  a  substance  with  an  ion  in  common  causes 
the  formation  and  separation  in  the  solid  state  of  the 
un-ionized  substance.  In  other  words,  the  term  5  still 
retains  its  constant  value,  and  consequently  the  con- 
stituent ions  of  the  substance  unite  to  form  more  of  the 
un-ionized  portion,  which,  since  the  solution  is  already 
saturated  with  it,  separates  out  as  solid.  This  has  been 
found  to  be  true  by  experiment,  but  only  true  for  those 
substances  which  are  difficultly  soluble.  The  effect  may 
be  observed  most  easily  by  dissolving  the  difficultly 

soluble  substance  in  a  solution  of  the  salt  with  an  ion 

» 

in  common;  but  it  can  also  be  attained  by  adding  to 
the  saturated  water  solution  of  the  substance  a  strong 
solution  of  the  salt,  when  a  precipitation  of  the  sub- 
stance, usually  in  the  crystalline  state,  will  be  observed. 
Thus  if  we  add  to  one  portion  of  a  saturated  solution  of 
silver  acetate  a  strong  solution  of  sodium  acetate  contain- 
ing x  moles  of  CH3COO'  ions,  and  the  same  amount  of 


ELEMENTS  OP  PHYSICAL  CHEMISTRY. 

a  solution  of  silver  nitrate  containing  x  moles  of  Ag*  to 
the  liter  to  another  equal  portion,  we  observe  an  equal 
precipitation  of  solid  silver  acetate  in  the  two  solutions. 

Although  all  this  is  true,  as  far  as  the  precipitation 
is  concerned  for  all  saturated  solutions,  it  is  only  for  the 
difficultly  soluble  substances  that  the  quantitative  rela- 
tions are  found  to  hold. 

The  examples  below  will  serve  to  show  how  the  solu- 
bility product  of  a  substance  can  be  found,  and  how 
when  once  found  can  be  employed  to  foresee  the  solubility 
of  the  substance  in  a  solution  already  containing  a  com- 
mon ion. 

Silver  bromate  is  soluble  at  25°  to  the  extent  of  0.0081 
moles  per  liter.  Assuming  'the  ionization  in  this  state  to 
be  practically  complete,  and  it  certainly  is  nearly  so,  the 
concentration  of  the  ions  Ag*  and  BrO3'  will  be  the  same, 
and  equal  each  to  0.0081  mole  per  liter.  The  solubility 
product  at  this  temperature,  then,  will  be 

(0.008 1 )  (0.008 1 )  =^AgBr03. 

The  solubility  in  a  solution  of  silver  nitrate  containing 
o.i  mole  of  Ag*  ions  (or  in  potassium  bromate  containing 
o.i  mole  of  BrOs'  ions)  can  be  found  by  aid  of  the  rela- 
tion 

(0.0081  )2  =  (0.0081  +.1  —y)  (0.008 1  —y), 


CHEMICAL  CHANGE.  309 

and  is  equal  to  (0.0081  —y)t  for  that  is  the  concentration 
of  Ag*  and  BrOa'  ions  now  existing  in  the  solution,  and 
coming  from  the  salt;  the  amount  o.i  of  one  being  -due 
to  the  other  salt,  and  y  being  the  AgBrO3  remaining  un- 
dissolved  owing  to  the  presence  of  this  o.i  mole  of  Ag*  or 
Br03'. 

This  is  true  for  all  binary  salts  when  they  can  be  as- 
sumed to  be  completely  ionized,  or  practically  so. 

Where  the  substance  dissociates  into  more  than  two 
ions  and  can  be  assumed  to  be  completely  ionized,  the 
relation  is  quite  similar.  Suppose  the  salt  to  dissociate 
according  to  the  scheme 


As  solubility  product  we  shall  have,  if  c  is  the  solubility 
of  the  completely  ionized  salt  MA3, 


or 


for  we  must  have  three  times  the  number  of  moles  per 
liter  of  A'  as  we  have  of  M'"  according  to  the  chemical 
equation,  i.e., 

C  =  CM-     and      <;  =  c/. 


310  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

In  case  of  solution  in  the  presence  of  o.i  mole  of  one 
of  the  ions,  we  have,  then, 


or 

(CM--*)  (£A'+o.i-3*)3 

from  which  it  is  apparent  that  the  effect  of  equal  ad- 
dition is  not  the  same  for  the  two  ions,  i.e.,  that  x  and 
y,  the  decreases  in  the  solubility,  are  not  equal. 

In  the  case  the  substance  is  not  completely  ionized,  the 
solubility  product  is  not  so  directly  related  to  the  solu- 
bility of  the  substance  as  in  the  above  cases,  i.e.  to  the 
square  in  one  case  and  twenty-seven  times  the  fourth 
power  in  the  other.  Consider  the  case  of  uric  acid, 
which,  at  25°,  is  soluble  to  0.0001506  mole  per  liter,  and 
is  ionized  in  that  condition  to  9.5%  into  H*  and  the 
negative  radical  which  we  shall  designate  as  U.  The 
solubility  product  here  is  naturally 
(0.0001506X0.095)  (0.0001506X0.095) 

=  SHU  =KHu  (0.0001506X0.905). 

The  solubility  of  uric  acid  in  a  molar  solution  of  hydro- 
chloric acid,  for  which  01  =  0.78  (i.e.,  H'  =0.78,  Cl'  =  o.78), 
is  to  be  found  in  the  following  way  : 

(0.0001506X0.095+0.78  —  #)  (0.0001506X0.09  5  —  x) 

—  (o.oooi  506  X  0.095)2, 


CHEMICAL   CHANGE.  311 

where  (0.0001506X0.095—*)  represents  the  present  con- 
centration of  H*  and  U'  from  the  uric  acid,  and  its 
total  solubility  in  the  hydrochloric  acid  solution  is 
(0.0001506X0.905)  +  (0.0001506X0.095—  x),  i.e.,  is  equal 
to  the  sum  of  that  which  is  un-ionized  and  that  which  is 
ionized. 

Just  as  for  organic  acids  in  general,  an  infinite  excess 
of  one  of  the  ions  will  cause  the  ionization  of  the  sub- 
stance to  become  zero.  //  is  to  be  observed  here,  however, 
that  this  excess  will  only  cause  the  solubility  to  become  zero 
in  the  case  that  the  ionization  is  compete.  In  the  case  of 
uric  acid,  an  infinite  amount  of  H"  or  U'  at  best  can  only 
reduce  the  solubility  by  9.5%,  the  remaining  90.5% 
being  un-ionized  and  not  affected  at  that  temperature  by 
any  addition  of  substance  which  does  not  react  chem- 
ically with  it. 

That  a  substance  is  always  decreased  in  solubility  by 
the  addition  of  a  substance  with  an  ion  in  common  is 
not  true,  as  the  well-known  behavior  of  silver  cyanide  m 
potassium  cyanide  will  show.  In  all  such  cases,  however, 
the  equilibrium  which  has  previously  existed  is  altered 
in  some  way,  so  that  the  relations  are  not  the  same. 
These  cases  are  usually  characterized  by  the  formation  of 
a  complex  ion ,  the  product  of  which  is  exceeded.  The 
removal  of  the  ions  to  form  this  complex  ion  disturbs 
the  equilibrium  of  the  difficultly  soluble  salt;  the  un-ionized 
portion  ionizes  further,  and  its  loss  is  replaced  by  the 


312  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

solid  phase.  This  process  continues,  dissolving  new  salt, 
until  equilibrium  is  attained,  i.e.,  until  the  solubility 
product,  whatever  it  may  be,  is  just  satisfied,  when 
solution  ceases.  The  ionization  of  silver  potassium 
cyanide  takes  place  almost  completely  according  to  the 
scheme 


but  it  has  been  found  by  Morgan  (Zeit.  f.  phys.  Chem.,  17, 
513-535,  1895)  that  in  a  0.05  molar  solution  we  have  Ag* 
ions  to  the  extent  of  3.5Xio~n  and  CN'  to  2.76Xio~3 
moles  per  liter. 

Knowing,  the  concentration  of  the  metal  ions,  for 
example  (which  can  be  determined  by  methods  given 
in  the  next  chapter),  in  the  complex  salt  solution  and  in 
a  water  solution  of  the  difficultly  soluble  salt,  we  can  fore- 
see the  behavior  of  that  salt  when  in  a  solution  of  a  salt 
which  might  dissolve  it  to  form  a  complex  solution  of 
that  strength.  In  general,  i.e.,  when  the  concentration 
oj  metal  ions  in  a  water  solution  of  salt  is  greater  than  that 
of  a  water  solution  of  a  complex  salt,  the  simple  salt  will 
dissolve  in  any  solution  which  will  produce  the  complex 
salt  in  this  concentration.  If  the  concentration  of  metal 
ions  is  smaller,  the  solid  will  not  dissolve  to  any  greater 
extent  than  it  does  in  pure  water,  for  the  ionic  product 
of  the  complex  ion  cannot  be  exceeded. 

By  this  law  it  is  possible  to  find  the  relative  solubility 


CHEMICAL    CHANGE.  313 

of  salts  of  the  same  metal  in  water.  Thus  silver  sulphide 
is  the  only  silver  salt  which  will  not  dissolve  in  potassium 
cyanide  solutions;  in  other  words,  is  the  most  insoluble 
salt  of  silver,  and  contains  fewer  ions  of  Ag*  than  exist 
even  in  a  solution  of  silver  potassium  cyanide,  such  as 
that  given  above. 

It  is  not  only  for  substances  in  solution  that  we  find 
this  constancy  of  the  product  of  the  concentrations  of 
the  ions,  for  it  also  exists  in  our  usual  solvent,  water, 
where  the  ionized  portion  is  so  small  that  the  un-ionized 
portion  may  be  considered  as  constant,  i.e.,  i—  a  =  i. 
Expressing  the  concentrations  of  H*  and  OH'  ions  in 
a  liter  of  water  by  Ci  and  c2,  and  the  un-ionized  portion, 

which  is  practically  i  liter,  i.e.,  —  0-^55-S  moles,  by  c, 

we  have 

5  =  *H2o  =  constant. 


The  values  of  Ci  =  c2  =  H'    (  =  OH')  ions  in  water  at 
various  temperatures  is  as  follows: 


Temp. 

CiXio7. 
Moles  per  Liter. 

Temp. 

C!  X  10*. 

Moles  per  Liter. 

0° 

o-35 

34° 

1-47 

10° 

0.56 

50° 

2.48 

1  8° 

0.80 

85°-5 

6.20 

'5° 

1.09 

100° 

8.50 

The  ionic  products  (we  can  hardly  call  them  solubility 
products),  then,  are  as  follows: 


3H  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


*o°=  (0.35X10-7)2,  0-34°  =( 

S10°  =  (0.56XIO-7)2,        550°  =(2.48X10-7)2, 
5i8°  =  (0.80  X  10-7)2,       585.5°  =  (6.2  X  10-7)2, 
*25°  "  (1  .09  X  TO-7)2,        5100°  =  (8.5  X  10-7)2, 

where  the  value  of  s,  in  each  case,  is  equal  to  55.5  times 

a2 
the  value  of  K=-(  -  y~,  V  being  0.0018  liter,  i.e.,  the 

volume   occupied   by  the  formula  weight,  18  grams,  of 
water. 

Knowing  the  solubility  products  of  two  substances 
with  an  ion  in  common,  it  is  possible  to  find  how  much 
of  each  will  dissolve  when  they  are  exposed  together 
to  the  action  %of  a  solvent;  and  this,  of  course,  may  be 
expanded  to  three  or  more  substances  together. 

Assume  we  have  the  two  completely  ionized,  diffi- 
cultly soluble  salts  MA  and  MAi,  with  the  ion  M*  in 
common,  and  that  they  are  dissolved  simultaneously 
in  water.  Call  the  amount  of  MA  which  dissolves  x, 
and  the  amount  of  MA  i  y.  In  the  solution  then  we  must 
have  x  +  y  moles  of  M"  ions,  x  of  A'  and  y  of  A\  \  and, 
if  5  is  the  solubility  product  of  MA  and  s\  that  of 
the  relations  must  be 


so  that   by  solving  the  simultaneous  equations  we  can 
find  x  and  y. 


CHEMICAL   CHANGE.  315 

An  example  of  this  is  given  by  dissolving  thallium 
chloride  and  sulphocyanate  together.  The  solubilities 
in  water,  each  for  itself,  are  TlCl  =  o.oi6i  and  T1SCN 
=  0.0149.  Assuming  complete  ionization,  the  solu- 
bility products  are  respectively  (o.oi6i)2  and  (o.oi49)2 
and  if  x  represents  the  amount  of  chloride  and  y  that 
of  sulphocyanate  dissolving  from  the  mixture,  we  have 
TT9  x=CY,  and  ;y  =  SCN',  and 


x(x+y)  =  (o.oi6i)2, 
y(x+y)  =  (0.0149)2, 

from  which  we  find  x=  0.0118  and  y  =  o.oioi,  while 
the  values  #=0.0119  and  y  =  0.0107  are  found  by 
experiment. 

It  will  be  observed  that  in  the  above  examples,  except 
the  last,  we  have  tacitly  assumed  that  the  dissociation 
of  the  added  salt,  with  an  ion  in  common,  is  not  influenced 
by  the  ions  of  the  difficultly  soluble  salt.  As  a  rule  this 
is  true,  for  the  substances  are  so  insoluble  that  their 
effect  is  infinitesimal;  in  the  last  example,  this  effect 
has  been  allowed  for,  however,  and  will  show  the  method 
of  treating  such  cases. 

In  general,  then,  we  can  conclude  for  difficultly  soluble 
salts  (and  for  complex  ions]  that  they  are  precipitated 
(formed)  when  the  product  of  the  concentrations  of  the  ions 
composing  them  exceeds  the  solubility  (ionic)  product. 


316  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Although  this  law  holds  in  general  for  difficultly  soluble 
salts,  isolated  cases  are  to  be  found  where  the  un-ionized 
portion  does  not  remain  rigidly  constant,  after  the  addi- 
tion of  an  ion  in  common;  and,  to  a  smaller  extent,  a 
slight  variation  is  sometimes  observed  in  the  solubility 
product.  Since  these  cases  are  very  few,  and  are  usually 
observed  for  the  more  soluble  salts,  it  would  seem  prob- 
able that  they  are  due  to  secondary  reactions  not  yet 
recognized,  or  to  others  not  properly  accounted  for. 

75.  Hydrolytic  dissociation.  Hydrolysis.  —  Hydrolysis 
is  the  process  taking  place  in  the  water  solution  of  a 
salt,  which  causes  the  solution  to  appear  alkaline  or  acid, 
or  results  in  a  neutral  equilibrium  according  to  the 

scheme 

=  MOH+HA. 


If  the  acid  formed  is  insoluble  or  un-ionized,  the  base 
being  ionized,  the  reaction  will  be  alkaline  (OH'  ions). 
When  the  base  is  insoluble  or  un-ionized,  and  the  acid 
ionized,  the  reaction  is  acid  (H*  ions).  And  finally, 
if  both  acid  and  base  are  insoluble  or  un-ionized,  the 
salt  will  be  completely  transformed  into  base  and  acid, 
and,  as  there  will  remain  no  excess  of  either  H*  or  OH' 
ions,  the  reaction  will  be  neutral.  In  other  words,  then, 
hydrolysis  is  due  to  the  removal  of  either  H*  or  OH'  ions 
(or  both)  from  the  water  by  the  A'  or  M*  ions  of  the 
salt,  to  form  un-ionized  or  insoluble  substances,  and  this 
continually  causes  more  water  to  ionize,  i.e.,  to  react 
with  the  salt, 


CHEMICAL   CHANGE.  317 

Examples  of  this  process  are  most  common.  For 
instance,  all  mercury,  copper,  zinc,  etc.,  salts  are  acid, 
for  an  un-ionized  basic  substance  is  formed  by  the  reac- 
tion leaving  free,  ionized  acid;  and  potassium  cyanide 
is  alkaline,  owing  to  the  formation  of  un-ionized  hydro- 
cyanic acid  and  ionized  potassium  hydrate.  The  most 
striking  example  of  this  process,  perhaps,  is  the  precipi- 
tation of  bismuth  oxychloride  when  water  is  added  to  a 
hydrochloric  acid  solution  of  the  chloride,  but  the  basic 
acetate  separation  of  iron  (see  below)  is  just  as  charac- 
teristic, although  apparently  not  so  direct. 

Since  we  know  the  conditions  under  which  insoluble 
or  un-ionized  substances  will  form,  i.e.,  by  the  exceeding 
of  their  solubility  products  or  analogous  values,  it  is 
possible  to  find  the  conditions  necessary  to  produce  a 
hydrolytic  dissociation,  and  also  to  find  the  relations 
governing  the  equilibrium  finally  attained  as  the  result 
of  the  process. 

We  recognize  at  once  that  if  the  product  of  the  con- 
centrations of  .M"  and  OH'  ions  is  larger  than  that  which 
can  exist  in  pure  water,  un-ionized  substance  must  form. 
By  this  formation,  however,  the  equilibrium  of  H*  and 
OH'  ions  will  be  disturbed,  and  a  further  ionization  of 
water  must  take  place,  until  at  length  the  ionic  product 
is  just  attained.  If  the  H*  and  A'  ions  at  this  point  do 
not  unite  to  form  un-ionized  acid,  the  further  ionization 
of  water  will  be  unlike  what  it  would  be  in  the  absence 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


of  this  excess  of  H*  ions,  for,  since  CH-XCQH'  must  at 
the  same  time  be  equal  to  sH2o>  we  can  only  have  —  - 

moles  per  liter  of  OH'  ions  present,  when  CH-  is  the 
total  concentration  of  H*  ions  at  that  time. 

The  process  due  to  the  formation  of  un-ionized  or 
insoluble  acid,  when  no  un-ionized  base  is  formed,  or 
forms  but  slightly,  is  exactly  analogous  to  the  above. 
In  both  cases  water  is  decomposed,  owing  to  the  removal 
of  one  of  its  ions,  and  the  further  ionization  of  water  and 
formation  of  the  insoluble  or  un-ionized  base  or  acid 
continues  until  the  equations  for  equilibrium  are  fulfilled. 

For  the  sake  of  simplicity  we  shall  consider  separately 
the  cases  that  the  reaction  is  caused  by  the  base,  or  by 
the  acid. 

Case  I.  The  process  is  due  only  to  the  formation  of  base. 
Here  it  is  obvious  that 


or 


where  the  terms  c  refer  to  the  ionic  concentrations.  Cal- 
ling c  the  original  concentration  of  salt,  and  d$  the  ioni- 
zation of  the  salt,  the  concentration  of  OH'  ions  in 
water  at  25°  being  1.09  Xio~7  moles  per  liter,  we  shall 
have 

-  or 


After  equilibrium  has  been  established,  i.e.;  when  the 


CHEMICAL   CHANGE.  3*9 

degree  of  hydrolytic  dissociation  is  a,  d&  being  the  dis- 
sociation of  the  acid  formed,  we  must  have 


and 

(470)     (orig.  M'  —  loss  of 

IV   'Lvll       1    A. 

=  ^MOH  XMOH  formed, 
or 

(476)  d$c(i-a) 


where,  if  the  base  has  a  solubility  product,  it  is  to  be  used 
in  place  of  the  terms  on  the  right,  and  the  value  SHaO 
varies  with  the  temperature,  having  the  value  (1.09  X  io~7)2 
at  25°. 

Case  II.     The  process  is  due  only  to  the  formation  of 
acid.     Here 


X  £H-  >      HA^HA      or 


Just  as  above,  since  CH-  =  i.o9Xio~7  at  25°,  we  shall 
have,  when  c  is  the  concentration  of  salt,  and  d$  its  ioni- 
zation, 


or 


320  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

If  ^B  is  the  ionization  of  the  base,  c  the  original  con- 
centration of  "salt,  and  a  its  hydrolytic  dissociation,  then 


total  H-- 


total  OH" 
and 


formed, 
or 

(486)  dsc(i—at 


And,  here  again,  the  solubility  product  may  be  used  on 
the  right,  if  the  acid  is  difficultly  soluble. 

The  formulas  above,  in  both  cases,  may  also  be  written 
in  another  form,  which,  although  it  does  not  illustrate 
so  well  the  principles  involved,  is  more  useful  in  many 
ways.  From  (47  ft),  by  transformation,  we  obtain 


Cds(l  -a 


or 


fr    ™)v  y- 

(i-a)v   as 

a 


CHEMICAL  CHANGE.  3*1 

and,  from  (486),  in  the  same  way, 

K\  <*2          <%       „ 

'-*hyd- 


In  dilute  solutions  where  d$,  d^  and  JB  rnay  be  regarded 
as  unity,  (490)  and  (496)  are  simplified  to  the  form 


yd.      /  vr   ~  v 

(i-a)F    AHA 

We  have  the  following  law,  then,  governing  hydrolysis. 
The  expression  for  the  hydrolytic  dissociation,  of  a  salt 

a2 
in  water,  ,   _     v,  is  equal,  when  due  to  the  formation  of 

base  (acid),  to  the  ionic  product  of  water  multiplied  by 
the  degree  of  ionization  of  the  salt  divided  by  the  ioni- 
zation  constant  of  the  base  (acid)>  multiplied  by  the  de- 
gree of  ionization  of  the  acid  (base). 

This  law  has  recently  been  much  used  to  determine 
from  the  experimentally  observed  hydrolysis  of  the  salt, 
the  ionization  constant  of  the  acid  or  base  formed. 

The  constant  of  hydrolytic  dissociation,  in  such  a  case, 
can  also  be  defined  (see  pp.  235  and  285),  when  d§,  d& 
or  JB  are  equal  to  i,  as  one-half  the  concentration,  in 
moles  per  liter,  at  which  the  salt  is  50%  hydrolyzed. 

And  if  a  is  small  enough  to  be  neglected  in  the  term 

a2 
i—  a,  /  _\y  =  K  is  also  reduced  (p.  290)  to 


322  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

in  other  words,  for  the  same  substance,  the  hydrolytic 
dissociation,  when  small,  is  proportional  to  the  square-  root 
of  the  dilution  of  the  salt,  i.e., 


aoc\/F    or    -y-, 


where  c,   the  reciprocal  of   V,   is  the  original  concen- 
tration of  the  salt  dissolved. 

Knowing  the  constant  for  hydrolytic  dissociation  it  is 
also  possible  to  calculate  the  degree  of  hydrolysis  at  any 
dilution  by  the  formula 


-si 


The  following  examples  will  serve  to  show  the  use 
which  may  be  made  of  the  above  relations. 

What  is  the  ionic  product  for  water  at  25°?  A  o.i 
molar  solution  of  sodium  acetate  is  0.008%  hydrolyzed; 
the  sodium  acetate  to  be  considered  as  completely  ionized, 
as  is  also  the  sodium  hydrate  formed,  and  the  ionization 
constant  of  acetic  acid  is  0.000018. 

Here 

CH3COOH  =  OH' =  0.00008X0.1  =0.000008 
and  since 


CHEMICAL  CHANGE.  323 

.00001  8  X  CcHaCOOH  =  ^H'  X  CAc>  , 

o.ooooi  8  X  0.000008 


=i.44Xio~8 


and 

Vi.  44  Xio~9X  0.00008 


What  is  the  hydrolysis  of  a  o.i  molar  solution  of  potas- 
sium cyanide  (assu ming d$  =  i )  ?  K for  HCN  =  13X10" 10 
and  %2o  =  (i.o9Xio~7)2  at  25°. 

a2       _(i.Q9Xio~7)2 
Ahyd'~(i-a)F:      13 Xio-10   ' 

from  which,  when  d%=i  and  F=io,  #=0.967%. 


In  the  table  below  are  given  the  values  of      _   .„  for 

various  equilibria  in  which  but  i  mole  of  water  reacts 
with  the  substance. 


HYDROLYSIS  OF  HYDROCHLORIDES  AT  25°. 


Base. 
Anih'ne  

Per  Cent 
Hydrolysis  a 

•    2    7 

b                                                         j 

(i-a)F' 
2    2SXIO-5 

[onization  Constant 
of  Free  Base. 

c    -j    XlO~10 

0-Toluidine        .  .  .  . 

7  O 

I   62X10^"* 

7    -I    V  IO—  ll 

nt-       '  ' 

^  6 

2    O     X  IQ—  10 

1>-       " 

i  8 

I  05X10—* 

I   13X10-° 

o-Nitroaniline        .  . 

98  6 

2    I 

e   6   Xlo~15 

m-          "           .... 
#-          " 

26.6 

70    6 

3-oiXio-3 
o  (;8Xio-2 

4.0  Xio~12 
i  24X10"  13 

Aroinoazobenzene 

.    .   18  i 

I    2!CXlO-3 

o  c   Xio—10 

Urea. 

0.781 

y-o    /v*w 

i.C    Xio-14 

324  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Thus  far  we  have  only  considered  that  one  mole  of 
water  reacts  with  the  salt;  in  other  words,  we  have  only 
employed  salts  containing  monovalent  elements.  In 
case  the  reaction  involves  more  than  one  mole  of  water 
the  treatment  is  the  same  as  for  the  law  of  mass  action 
in  general.  It  must  be  said,  however,  that  such  cases, 
so  far  as  we  know  at  present,  are  not  at  all  common; 
the  salt  often  reacting  with  but  one  mole  of  water  .  to 
form  a  basic  salt  which  still  retains  some  of  the  original 
element.  Wherever  the  relation  of  hydrolysis  to  con- 
centration is  that  given  on  page  316,  this  is  so.  There 
is  one  reaction,  however,  which  gives  a  good  constant 
assuming  two  moles  of  water  to  react,  and  we  shall  con- 
sider it  to  show  how  such  relations  are  to  be  treated.  The 
reaction  is 

A1C13  +  2H2O  =  A1(OH)2C1  +  2HC1, 

which  has  been  investigated  by  Kullgren  (Om  metalls- 
alters  hydrolys,  page  108.  Dissertation.  Stockholm, 
1904).  We  have,  then,  similarly  to  (476),  where  c  is 
the  molar  concentration  of  AlCls  in  solution,  and  a  is  the 
fraction  of  it  hydrolyzed,  ds  being  the  ionization  of  Aids, 
and  JA  that  of  the  HC1,  of  which  20.0  is  formed, 


It  is  impossible  to  use  this  formula  in  calculations,  how- 


CHEMICAL  CHANGE  325 

ever,  for  as  yet  we  know  nothing  of  ^Ai(OH)2ci-  By  using 

the  formula  in  the  other  form,  analogous  to  (490),  we 
obtain 


_ 

J 


and  from  this  we  can  calculate  the  ionization  constant 
of  A1(OH)2C1,  when  a  is  known,  or  dispense  entirely 
with  it,  i.e.,  using  the  K^A.  so  determined  for  the  cal- 
culation of  other  values.  It  will  be  observed  here  that  a, 
instead  of  being  proportional  to  \/V  as  it  is  for  the  reac- 
tion with  i  mole  of  water,  is  proportional  to  3/V2.  The 
following  results  will  show  how  well  this  equilibrium 
follows  the  above  law,  and  how  it  is  possible  to  find  the 
ionization  constant  by  aid  of  the  hydrolytic  dissociation, 
knowing  the  ionic  product  for  water  at  that  temperature. 

HYDROLYSIS  OF  A1C18  AT  100°  C. 
A1C13+3H20  =  A1(OH)2C1+2H- 

•tr  d  A  Ac*  a8 


¥ 

a 

X*. 

o 

(i-a)T/2 

(i-a)V»-~ 

96 

o  .  1488 

0.966 

0.76 

420XIO-' 

5l6XlO-' 

384 

0.3629 

0.977 

0.85 

509X10-' 

57lXlO-' 

1536 

0.7142 

i 

0.91 

54IXIO-' 

594X  io-» 

Average,  -Khyd.  =-  560  X  io~  * 

The  concentrations  of  base  in  the  three  cases  are  0.00155, 
0.000945  and  0.000465,  respectively,  the  acid  concen- 
trations being  twice  these  values.  The  average  value  of 


326  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

#hyd.  in  the  last   column    may  be    used    to    determine 

^Ai(OH)2ci>    for  H2°     =4—  -yFT2.  -A   We  obtain 

AA1(OH)2C1         (1-ajK^       (1$ 

in    this    way   the   value   ^A1(oH)2ci=2-33XIO~195    where 
the  ionization,  presumably,  gives  Aid"  and  2OH'. 

The  formation  of  a  substance  containing  OH  as  well 
as  the  original  negative  element  is  very  common.  In 
the  case  of  the  chloride  of  bismuth  mentioned  above  the 
substance  separating  out  by  hydrolysis  is  not  the  pure 
hydrated  oxide,  but  an  oxychloride;  but  apparently 
this  is  not  true  in  the  case  of  the  hydrolysis  of  ferric 
chloride.  In  this  case  the  reaction  is  not 

FeCl3  +  3H20  -  Fe(OH)3  +  3H'  +  3C1  , 

but,  according  to  Goodwin  (Zeit.  f.  phys.  Chem.,  21,  i, 
1896,  and  Phys.  Rev.,  n,  193,  1900),  must  rather  be 


where  the  FeOH"  is  colloidal. 

In  certain  other  cases  it  has  been  found  that  hydrolytic 
dissociation  takes  place  in  stages,  i.e.,  first  i  mole  of 
water  reacts,  then  another,  etc.  It  is  quite  certain,  how- 
ever, that  this  does  not  occur  at  the  dilutions  above  of 
Aids,  for  if  it  did,  the  formula  used  would  not  give  a 
constant  value,  hence  in  this  one  case  between  these 
limits  of  dilution  2  moles  of  water  react  with  i  of  salt. 

T1(NO3)3,  according  to  Spencer  and  Abegg  (Zeit.  f. 


CHEMICAL   CHANGE  327 

anorg.  Chem.,  44,  397,  1905),  however,  seems  to  react 
directly  with  3  moles  of  water,  T1(OH)3  having  a  solu- 
bility equal  to  jo"13-58  moles  per  liter,  i.e.,  j=io~52-896, 
but  as  yet  this  is  the  only  case  known. 

We  must  now  consider,  very  briefly,  the  methods  by 
which  the  degree  of  hydrolytic  dissociation  can  be  experi- 
mentally determined.  It  is  obvious  from  what  has 
already  been  said,  that  the  determination  of  the  concen- 
tration of  any  one  of  the  reacting  constituents,  together 
with  the  chemical  reaction,  will  give  us  a  complete  view 
of  the  equilibrium. 

The  general  methods  so  far  used  for  this  purpose 
are  as  follows: 

(1)  The  free  acid  (H")  or  free  base  (OHO  is  measured 
by  observations  of  the  velocity  of  the  inversion  of  sugar 
at   100°,  or  the  hydrolysis  of   esters,  as  was   described 
above  (pp.  269,   270). 

(2)  The  determination  of  the  ionized  portion  by  means 
of  conductivity  observations. 

For  details  as  to  these  methods  see  Ley  (Zeit.  f.  phys. 
Chem.,  30, 193, 1899),  Kullgren  (1.  c.),  and  Goodwin  (1.  c.). 

When  one  of  the  active  constituents  is  colored,  the 
reaction  may  also  be  followed  by  aid  of  spectro-photo- 
metric  observations,  as  shown  by  Moore  (Phys.  Rev.,  12, 
151-176,  1900),  but,  naturally,  this  method  is  much 
limited  in  its  applicability. 

One  method,  which  can  be  used  for  salts  of  weak  acids 


328  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

with  strong  bases,  or  salts  of  weak  bases  with  strong 
acids,  has  been  suggested  by  Farmer  (Trans.  Chem. 
Soc.,  79,  863,  1901,  and  ibid.,  85,  1713,  1904),  which, 
owing  to  the  importance  of  the  principle  involved,  is 
briefly  considered  below.  The  method  is  based  upon 
the  coefficient  of  distribution  of  a  substance  between 
water  and  another  solvent,  benzene  (p.  188).  Thus, 
hydroxyazobenzene  has  a  coefficient  of  distribution  be- 
tween water  and  benzene  equal  to  539,  i.e.,  benzene 
always  takes  up  539  times  as  much  hydroxyazobenzene 
as  the  water,  when  the  two  solvents  are  present  in  equal- 
volumes.  If  the  two  solvents  are  present  in  unequal 
quantity,  say  i  liter  of  water  to  q  liters  of  benzene,  the 
hydroxyazobenzene  in  the  water  will  be  distributed 
between  them  in  the  ratio  1:539?. 

By  shaking  a  water  solution  of  the  barium  salt  of 
hydroxyazobenzene  with  benzene,  then,  the  free  hydroxy- 
azobenzene, if  it  be  formed  by  hydrolysis,  will  be  partially 
extracted  by  the  benzene.  Finding  the  amount  of  this 

present  in  the  benzene  solution,  multiplying  it  by  -    — , 

we  find  what  is  left  in  the  aqueous  solution.  The  sum 
of  these  two  quantities,  then,  is  the  concentration  of  free 
hydroxyazobenzene  which  has  been  formed  as  the  result 
of  hydrolysis,  and,  knowing  the  amount  of  salt  initially 
present,  the  degree  of  hydrolytic  dissociation  is  easily 
calculated.  By  this  method,  for  numerous  dilutions  of 


CHEMICAL   CHANGE.  329 

a2 
the  barium  salt,  the  formula  K=-r-— — r~  was  found  to 

give  a  constant  value  for  K,  which  at  25°  is  equal  tc 
24.3X10-7. 

This  method  was  also  applied  to  the  hydrolysis  of  the 
hydrochlorides  of  weak  bases,  as  aniline,  etc.  (where 
the  coefficient  of  distribution  of  the  free  base  is  deter- 
mined), with  very  satisfactory  results.  The  values  in  the 
table  on  page  323  were  found  in  this  manner. 

In  all  such  determinations  constancy  of  temperature  is 
o]  paramount,  importance,  for  hydrolytic  dissociation,  as 
will  have  been  observed  from  the  foregoing,  is  largely  in- 
fluenced by  the  temperature.  This  is  due  not  only  to  the 
increased  ionization  of  water  (p.  313)  with  the  temper- 
ature, but  also  to  the  decrease  in  the  ionization  constants 
of  acids  and  bases.  Thus  for  acetic  acid  #i8°  =  i8.3X 
io~6,  #ioo°=n.4Xio-6,  ^i56°=5.6Xio~6  and  K2^°  = 
1.9  X  i o~6,.  while  for  ammonium  hydrate  ^i8°=i7.iX 
io~6,  Kioo°=i4Xio~6,  and  J£i56°  =  6.6Xio~6. 

76.  Determination  of  the  ionization  constant  from  ob- 
servations of  increased  solubility. — A  very  ingenious 
and  accurate  method  for  the  determination  of  the  ioniza- 
tion constant  of  an  acid  or  a  base,  from  its  -increased 
solubility  in  a  base  or  an  acid,  with  a  known  ionization 
constant,  is  given  by  Lowenherz  (Zeit.  f.  phys.  Chem., 
I5>  385?  1898).  Either  the  acid  or  base  to  be  deter- 
mined must  be  difficultly  soluble  in  water,  i.e.,  just  those 


330  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

conditions  are  advantageous  which  in  the  usual  methods 
are  sources  of  trouble. 

The  general  form  of  the  equilibrium  arising  by  the 
neutralization  of  an  acid  with  a  base  may  be  written 
as  follows: 


+iA'+gM'+hOH'. 

Calling  the  total  amount  of  acid  present  2  A,  and  the 
total  base  IM  ,  we  have  the  following  conditions  existing 
at  equilibrium: 

(1)  A'  =  JA-HA-MA. 

(2)  A/1=  M'+H'—  OH'.     The  sum  of  the  positive  ions 

is  equal  to  the  sum  of  the  negative. 


(3)  M' 

(4)  M'=A'  +  OH'-H\ 


(5) 

KacidHA 

XX    —             .  ^         . 

(6) 

=  OH'" 

f*\ 

nTT,    KbaseMOH 

(8) 


CHEMICAL   CHANGE.  33 l 

(9)  MA  =  (i  -  O:MA)  total  MA. 

(10)  MOH  =  (i  -«MOH)  total  MOH. 

(ICKZ)  MOH,  when  difficultly  soluble  is  the  same  as  in  a 
saturated  water  solution. 

(n)  H2O  =  total  MA. 

(12)  HA=  (i  -«HA)  total  HA. 

(120)  HA,  when  difficultly  soluble  is  the  same  as  in  a 
saturated  water  solution. 

The  solubility  of  a  difficultly  soluble  acid,  for  example, 
will  be  greater  in  a  solution  of  a  base  than  in  pure  water, 
for  the  reaction  causes  the  removal  of  H*  ions  from 
the  acid,  and  this  necessitates  the  further  ionization  of 
the  un-ionized  portion,  and  the  solution  of  more  solid. 
The  difference  in  solubility  in  the  base  and  in  water  gives 
directly  the  amount  of  water  formed,  for  that  amount 
is  the  cause  of  the  reaction  which  increases  the  solubility. 
The  total  salt  formed  is  also  equal  to  this  difference, 
but  this,  unlike  the  water  formed,  may  be,  and  usually 
is,  ionized. 

The  sequence  in  which  the  equations  above  may  be  used 
in  any  individual  case  depends  entirely  upon  the  nature 
of  the  equilibrium.  The  first  thing  in  all  cases  is  to  find 
the  amount  of  salt  formed  andy  from  its  ionization  in 
this  dilution^  the  concentration  o\  its  ions.  The  following 


332  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

very  simple  examples,  however,  will  probably  do  more 
to  show  the  principles  involved  than  pages  of  explana- 
tions. 

Assume  we  have  an  acid,  soluble  to  o.ooi  mole  per 
liter,  the  constant  of  ionization  of  which  is  unknown, 
which  is  soluble  to  0.003  m°le  per  liter  in  a  o.i  molar 
solution  of  the  base  MOH.  The  base  is  90%  ionized, 
and  the  salt  which  is  formed,  has  a  value  of  a  equal  to 
98%  at  this  dilution.  What  is  the  constant  of  ionization 
of  the  acid? 


Here  we  must    find  ^HA^  -  •     Since    the    dif- 


ference  in  the  two  solubilities  is  0.002,  the  amount  of 
water  and  the  total  salt  formed  are  each  equal  to  0.002 
mole  per  liter.     We  have,  then, 
MA  =  (i  -a)  (total  MA) 

=  (1-0.98)  (0.002). 
M*  =  M'  of  MOH  -loss  of  M'  as  MA 

=  0.9(0.1)  —  (0,002)  (i  —0.98). 
OH'  =  OH'  of  MOH  -  loss  of  OH'  as  H2O 
=  0.9(0.1)  —  (0.002). 


H*  = 


SH2O 

OH' 

(1.09  Xio~7)2 


0.9  (o.i)  —(0.002)" 
=  H'+M*-OH' 


0.9(0.1)  -(0.002.) 

—[0.9  (o.i)  —  (0.002)] 


CHEMICAL  CHANGE.  333 

7+[°-9  (o.i)-  (0.002)  (i  -0.98)] 


(I.Q9XIQ-7)2 
o.c88 

—  (1—0.98)  (0.002). 
And  since 


we  can  find  its  value  by  substituting  the  values  given 
above,  i.e.,  H',  A',  and  HA. 

Proceeding  in  this  way,  and  only  in  this  way,  we  can 
find  the  concentrations  of  the  various  constituents  at 
equilibrium. 

The  sequence  of  determination  is  somewhat  different 
when  the  base  is  difficultly  soluble  and  Kbase  is  unknown, 
an  acid  solution  with  a  known  degree  of  ionization  serv- 
ing as  solvent. 

Assume  a  base,  which  is  soluble  to  o.ooi  mole  per 
liter,  to  be  soluble  in  an  acid  solution  (01  =  0.90)  to  the 
extent  0.003  m°le  Per  liter,  the  salt  being  ionized  to 


0.98%.    What  is  K=  for  the  base? 

\  ^MOH       / 

Here  again 

MA=  (o.oo2)(i  —0.98), 


334  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

but  now  we  must  first  determine  A'  instead  of  M*   as 
above,  i.e., 

A'  =  A'  of  HA  -  loss  of  A  as  MA 

=  0.9(0.  1  )  —  (O.002)(l  —  0.98). 

H*  =  H'  of  HA  -  loss  of  H*  as  H2O 
=  0.9(0.1)  —  (0.002). 


OH'  = 

±1 

(1.09  Xio~7)2 


0.9(0.1)  —  (0.002)" 
A'  +  OH'-H' 

[0.9(0.1)  -(0.002X1-0.98)]+    (1-09X10-7)2 

0.9(0.1)  -(0.002) 
-[0.9(0.1)  -(0.002)]. 

(1.00X10  7)2 


MOH=^M-M'-MA 
=  0 


(i.opXio-7)2"! 
.003-  [0.00196  +  -  -~-  --  J 

—  (0.002)  (l—0.98) 


and 


the  values  of  which  are  above,  viz  ,  M ',  OH',  and  MOH. 
In  case  the  acid  is  dibasic,  or  the  metal  of  the  base 
is  divalent,   the  appropriate  changes  may  be  made  in 
the  above  formulas  without  difficulty. 


CHEMICAL   CHANGE.  335 

In  all  cases,  however,  it  is  necessary  to  have  the  solvent 
(acid  or  base)  considerably  more  concentrated  than  the 
dissolved  substance  (base  or  acid). 

77.  Ionic  equilibria. — In  order  that  the  importance 
of  the  things  we  have  just  studied  may  be  more  clearly 
realized,  we  shall  now  consider  very  briefly  their  appli- 
cation to  a  few  questions  of  general  chemical  interest. 
Since  the  states  of  equilibrium  which  we  most  often  en- 
counter in  our  daily  experience  are  those  composed  of 
ionized  substances,  and  since  those  composed  of  un- 
ionized substances  are  comparatively  simple  and  easy 
to  determine,  we  shall  restrict  ourselves  here  to  the  con- 
sideration of  systems  containing  ionized  substances. 

The  simpler  cases  of  ionic  equilibria,  i.e.,  those  exist- 
ing in  simple  electrolytes,  have  already  been  considered 
above.  We  found  then  that  the  law  of  mass  action 
enables  us  to  foresee  the  equilibrium  in  systems  com- 
posed of  organic  acids  or  bases,  of  substances  which 
are  difficultly  soluble,  or  of  those  of  which  the  ionization 
is  very  slight.  But  for  other  systems,  characterized  by 
a  large  degree  of  ionization,  the  law  of  mass  action  is 
apparently  inapplicable,  and  must  be  replaced  by  certain 
empirical  relations. 

We  also  found  complications  to  arise  in  these  equi- 
libria, produced  by  the  further  dissociation  at  higher 
dilutions  of  one  of  the  two  ions  observed  at  lower  dilu- 
tions. This  is  the  case  with  the  poly-basic  organic 


336  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

acids,  which,  up  to  the  dilution  at  which  a  =  0.5,  behave 
as  though  monobasic,  only  showing  their  real  basicity 
above  this  dilution.  Even  in  such  a  case,  however,  the 
process  can  be  followed  or  foreseen  by  the  law  of  mass 
action,  each  form  of  ionization  leading  to  a  constant 
ratio  of  the  concentrations  of  the  constituents.  An  an- 
alogous case  with  a  strong  electrolyte  is  that  of  sulphuric 
acid,  which  ionizes  in  two  stages,  as  follows: 

H2SO4  =  H'+HSO4, 
HSO4'=H'+SO4". 

Here,  however,  we  cannot  follow  the  process  by  the 
law  of  mass  action,  and  the  first  stage  is  only  to.be  ob- 
served in  concentrated  solutions.  Indeed,  at  and  above 
a  dilution  of  i  mole  in  5  liters  no  trace  of  the  HSO/ 
ion  can  be  detected,  and  a  dilution  of  i  mole  in  1000- 
2000  liters  shows  practically  complete  ionization  into 
2H*  and  SO4".  The  behavior  of  salts  of  the  type  of 
BaCU  is  similar  to  this;  and  in  general  we  may  say 
that  the  more  dilute  the  solution  the  smaller  the  amount 
of  a  complex  ion  present. 

All  these  cases  of  equilibrium  are  comparatively  simple, 
however,  for  all  the  ionized  matter  present  arises  from 
the  one  original  substance  with  which  we  start.  The 
cases  we  are  now  to  consider,  i.e.,  those  equilibria  result- 
ing from  the  reaction  of  two  or  more  substances,  are 
somewhat  more  complicated  experimentally,  for  the 


CHEMICAL  CHANGE.  337 

ionized  matter  may  be  due  to  several  substances,  but 
theoretically  they  are  to  be  treated  just  as  the  others. 

Before  considering  the  specific  cases  of  equilibrium 
which  have  been  observed,  it  will  be  well  to  review  very 
briefly  the  various  methods  for  the  determination  of 
ionization  and  molecular  weight  in  solution,  in  order 
that  the  physical-chemical  analysis  of  such  systems  may 
be  quite  clear.  In  those  electrolytes  ionizing  into  two 
portions  measurements  of  the  electrical  conductivity  or 
freezing-point  depression  give  all  the  necessary  informa- 
tion, i.e.,  of  ionized,  as  well  as  of  un-ionized  matter. 
This  is  also  true  when  three  ions  are  formed  completely. 
When  one  of  the  three  ions  is  formed  to  a  smaller  extent 
than  the  others,  however,  as  is  the  case  with  the  second 
H*  ion  of  dibasic  organic  acids,  some  other  method,  in 
addition  to  the  ones  above,  must  be  employed  in  order 
to  show  its  concentration.  As  the  equilibrium  becomes 
still  more  complicated,  i.e.,  contains  other  ionized  and 
un-ionized  substances,  still  other  methods  must  also  be 
employed.  In  short,  the  physical-chemical  analysis  can 
only  be  accomplished  by  a  combination  of  two  or  more 
of  these  methods.  . 

Starting  with  a  given  solution  containing  various  ions 
and  un-ionized  substances,  by  the  application  of  all  or 
some  of  the  following  methods,  we  can,  of  course,  ascer- 
tain the  concentration  of  each  of  the  constituents  if  this  be 
necessary.  As  a  matter  of  fact,  however,  this  is  usually  un- 


ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

necessary,  for,  as  will  be  seen  from  page  330,  knowing  the 
values  of  certain  ones,  those  of  the  others  may  be  found 
by  aid  of  the  chemical  and  physical-chemical  relations. 

The  electrical  conductivity  of  a  solution  gives  directly 
the  total  concentration  o)  the  ions  which  it  contains;  while 
the  freezing-point,  boiling-point,  vapor  pressure,  or  osmotic 
pressure  give  the  total  number  oj  moles  per  liter,  i.e.,  of 
ionized  plus  un-ionized  matter.  These  are  the  general 
methods  applicable  to  all  kinds  of  ionized  or  un-ionized 
matter  and  their  application  to  an  unknown  solution  is 
obvious.  In  addition  to  these,  however,  there  are  certain 
other  special  methods,  applicable  to  certain  kinds  of 
ions  and  un-ionized  matter,  which  enable  us  to  make 
a  more  detailed  analysis  of  the  solution.  For  example, 
by  the  depression  of  the  solubility  of  a  substance  with 
an  ion  in  common,  the  concentration  of  that  ion  in  the 
solution  employed  as  the  solvent  may  be  calculated, 
and  with  great  accuracy.  The  nature  of  the  ions  in 
general  may  be  determined  by  migration  experiments, 
i.e.,  by  observing  the  changes  in  concentration  due  to 
the  passage  of  the  electric  current;  and  by  aid  of  electro- 
motive force  measurements  the  individual  concentrations 
of  many  kinds  of  ions  may  be  accurately  calculated. 
For  details  as  to  these  methods  see  Chapter  IX.  Con- 
centrations of  ionized  hydrogen  may  also  be  found  from 
the  speed  of  the  inversion  of  cane  sugar,  using  a  weak, 
known  solution  of  an  acid  as  a  standard;  and  OH'  ions 


CHEMICAL  CHANGE.  339 

can  be  determined  from  the  speed  of  saponification  of 
an  ester,  a  known  and  weak  solution  of  a  strong  base 
being  used  as  a  standard.  And,  finally,  un-ionized  ag- 
gregates can  be  determined  by  partition  or  distribution 
experiments,  i.e.,  by  finding  the  concentration  of  the 
substance  in  another,  immiscible  solvent,  after  this  has 
been  shaken  with  the  original  solution.  And  this,  it  is 
to  be  remembered,  is  not  restricted  to  any  one  un-ionized 
substance,  for  even  when  several  are  present  together  they 
behave  as  if  they  were  alone,  and  so  can  be  determined. 

By  these  methods,  then,  it  is  possible  for  us  to  find  the 
concentrations  of  the  various  substances  present  at 
equilibrium,  both  ionized  and  un-ionized,  and  thus  to 
define  exactly  the  conditions  necessary  and  sufficient 
for  the  retention  of  that  state. 

As  was  said  above,  we  are  now  to  consider  the  equilibria 
resulting  from  the  reaction  of  two  or  more  substances, 
but,  since  a  chemical  reaction  in  ionized  systems  depends 
largely  upon  the  formation  of  un-ionized  or  difficultly 
soluble  substances,  it  will  be  seen  at  once  that  the  law  of 
mass  action  is  applicable  to  these  systems.  For  such 
cases  where  it  is  inapplicable  we  have  no  guiding  principle 
at  present,  but  these  are  few,  so  far  as  we  know,  and  at 
any  rate  are  of  lesser  importance  in  ordinary  work. 

One  of  the  first  questions  arising  in  qualitative  analysis 
is — what  occurs  when  the  precipitation  of  Mg(OH)2 
by  ammonium  hydrate  is  prevented  by  the  presence 


340  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  ammonium  chloride?  The  older  theory  of  this 
influence  assumed  the  formation  of  a  complex  salt 
(MgCl4)(NH4)2,  which  is  not  decomposed  by  ammonium 
hydrate;  while  according  to  the  new  theory  it  is  due  to  the 
fact  that  the  ionization  of  the  ammonium  hydrate  is  so  de- 
creased by  the  presence  of  the  NH4*  ions  of  the  chloride 
that  the  solubility  product  of  Mg(OH)2  cannot  be  exceeded. 
This  question  was  first  investigated  by  Loven  (Zeit.  f. 
anorg.  Chem.,  37,  327,  1896).  If  the  action  is  due  to 
the  driving  back  of  the  ionization  of  NH4OH  by  the 
NH4*  of  NH4C1,  the  reaction  would  be 

MgQ2  +  2NH4OHrfVIg(QH)2  +  2NH4C1. 
And  in  every  case,  even  when  an  excess  of  NH4C1  is 
present,  we  must  have  the  following  relations  at  equilib- 
rium: 


or,  by  combination, 


G  J 

i.e.,  cMg- X I  — ^31  -  "TT"  =  *i  a  constant. 


Love*n  found  the  following  concentrations  at  equilib- 
rium, starting  with  different  salts  of  magnesium  and 
various  concentration  of  ammonia  and  ammonium 
chloride,  the  temperature  being  16-17°. 


CHZMlCAL  CHANGE.  34* 


Mg 

NH3 

NH4 

k 

0.0203 

0.0421 

O.OIO22 

o-34 

0.0281 

0.02027 

O.OO59 

o  33 

0.03762 

0.0189 

0.00655 

0.31 

o.  1084 

0.0499 

0.0286 

o-33 

Although  in  the  calculation  of  the  constant  k  the  ionic 
concentrations  of  Mg  and  NH4  should  be  used  in  the 
above,  complete  ionization  of  the  Mg  and  NH4  salts  are 
assumed,  and  the  total  amounts  employed.  As  will  be 
seen,  since  these  concentrations  were  determined  by 
ordinary  analytical  methods,  the  value  of  k  is  constant 
within  the  experimental  limits.  Loven's  own  formula 
was  expressed  somewhat  differently,  but  this  form  is 
used  here  so  that  it  may  be  in  accord  with  other  work 
of  the  same  kind. 

Muhs  (Dissertation,  Breslau,  1904)  studied  this  same 
equilibrium,  when  attained  from  the  other  direction, 
i.e.,  according  to  the  reaction 

Mg(OH)2  +  2NH4Cl<=±MgCl2  +  2NH4OH. 
Here,  just  as  above,  we  have 


and 

but,  since  no  NH4OH  was  added,  we  know  that 

hence 


2CMg" 

CNH;   J 


342  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

or 

c*  s 

—^ —  =  — \  9  2  =  constant, 
^NH;          4&i2 

i.e., 

1.5 

TCiir-  =  constant. 


At  29°  Muhs  found  the  following  results: 

Mg  NH4  Constant 

0.049  0.0771  0.141 

0.0638  0.106  0.152 

0.089  0.172  0.154 

0.108  0.25  0.140 

0.156  0.388  °-I59 

for  ammonium  chloride,  while  for  ammonium  nitrate  he 

found 

0.0495  0.076  0.145 

0.0833  0.049  0.131 

Here,  also,  the  ionization  is  assumed  to  be  complete, 
and  it  certainly  is  within  a  few  per  cent  of  being  so.  The 
values  of  the  constants  are  not  to  be  compared  here,  for 
unfortunately  the  temperatures  differ  widely,  which 
would  probably  exercise  a  great  influence  on  the  am- 
monium hydrate  solution,  as  well  as  upon  the  solubility, 
and  consequently  solubility  product,  of  Mg(OH)2. 

In  addition  to  these  independent  proofs  that  no  com- 
plex is  formed  in  such  solutions  we  also  have  another 
based  upon  entirely  different  principles  (Treadwell, 
Zeit.  f.  anorg.  Chem.,  37,  327,  1903).  The  molecular 
weight  of  a  solution  containing  the  chlorides  of  magne- 
sium and  ammonium  in  the  ratio  to  form  a  complex 


CHEMICAL  CHANGE  343 

salt,  if  it  did  exist,  can  be  found  from  the  following  data : 

0.1466  gram  of  MgCl2,  and  0.1647  gram  of  NH4C1 

(i.e.,  i  mole  of  MgCl2  to  2  of  NH4C1)  dissolved  in  20 

grams  of  water  cause  a  depression  of  the  freezing-point 

equal    to    o.°96o9.     From    this    the    average    molecular 

weight  of  the  substance  in  solution  is  found  to  be  29.97. 

Assuming  the  salts  to  be  present  as  a  mixture,  and  that 

they  are  practically  completely  ionized  (for  the  justification 

of  which  see  below),  the  average  molecular  weight  would 

be '- —  =  28.9,  for  a  mixture  of  i  mole  of  MgCl2  and 

2  moles  of  NH4C1  would  dissociate  into  seven  moles  of  ions. 
If  a  compound  were  formed  there  could  not  be  more  than 
three  ions  formed  at  most,  and  the  minimum  molecular 

202.32 

weight  in  solution  would  be —  =  67.44. 

«5 

Solutions  formed  by  each  salt  alone  in  this  dilution 
give  the  following  results: 

0.1466  gram  of  MgCl2  in  20  grams  of  water  depresses 
the  freezing-point  o.°3956,  from  which  ^  =  34.27,  instead 

MgCl2    95.26 
°f— 2=~3i.75. 

And  0.1647  gram  of  NH4C1  in  20  grams  of  H2O  gives 
a  freezing-point  depression  of  o°.55ii,  i.e.,  ^=27.75, 

e  NH4C1     53.52 
instead  of =^-^-=  26.74. 

These  results  show  that  the  ionization  of  each  alone 


344  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

at  this  dilution  is  nearly  complete,  and  certainly  the 
simultaneous  solution  cannot  change  the  ionization 
enough  to  significantly  change  the  result  for  the  average 
molecular  weight.  If  the  two  formed  a  compound,  the 
value  for  the  average  molecular  weight  could  not  under 
any  circumstances  be  as  low  as  29.97,  as  we  find  it.  This 
example  illustrates  very  well  the  difference  between  a 
complex  (or  compound)  salt  and  a  simple  mixture  of 
two  or  more  salts.  In  the  former  case  the  ionization 
is  changed,  i.e.,  a  complex  ion  is  formed;  in  the  latter 
the  ions  formed  are  the  same  as  those  which  exist  when 
the  substances  are  present  alone  in  the  solution. 

The  behavior  of  a  mixture  of  MnCU  and  NH4C1  can 
be  seen  from  the  following  results:  0.0968  gram  MnCl2 
and  0.0824  gram  NH4C1  (i.e.,  i  mole  to  2)  in  20  grams 
of  H2O  give  a  depression  of  the  freezing-point  equal 
to  o°.4797,  i.e.,  ^1=34.56.  Assuming  that  the  com- 
pound [MnCl4][NH4]2  is  ionized  into  three  ions,  it  could 
not  be  more,  the  average  molecular  weight,  just  as  above, 

2^2  06 

would  be  -       —=79.99,  while  a  mixture  of  the  two,  the 

o 

ionization  remaining  unchanged,  would  lead  to  —  = 

33-28. 

We  can  conclude,  then,  that  in  such  solutions,  neither 
system  forms  a  complex  salt,  and  that  the  behavior  0}  mag- 
nesium salts  with  ammonium  hydrate  simply  depends 


CHEMICAL   CHANGE.  345 

upon  the  concentration  oj  OH'  ions  present,  and  upon 
those  ions  which  can  alter  the  concentration  oj  these. 

One  of  the  questions  which  gave  much  trouble  before 
the  inception  of  the  electrolytic  theory  of  dissociation 
was  that  in  regard  to  the  sequence  of  separation,  when 
it  is  possible  for  two  salts  of  differing  solubility  to  be 
formed  in  a  solution.  According  to  the  older  theory 
it  was  assumed  that  the  more  insoluble  one  always  formed 
first,  and  could  be  separated  from  the  other  by  fractional 
precipitation.  This,  however,  is  found  not  to  be  in 
accord  with  the  experimental  facts,  which,  on  the  other 
hand,  are  perfectly  represented  by  the  law  of  mass  action. 
This  subject  has  been  investigated  by  Findlay  (Zeit.  f. 
phys.  Chem.,  34,  409,  1900)  in  the  case  of  the  reversible 

reaction 

PbSO4  -f  2NaI<=*PbI2  +  Na2SO4. 

solid         dissolved      solid          dissolved 

Applying  the  law  of  mass  action  to  this,  since  at  any  one 
temperature  the  concentrations  of  the  solids  in  solution 
are  constant,  we  have 


—  =  constant, 

CS04" 

for  the  reaction  may  also  be  written 


PbSO4  +  2l'    PbI2  +  SO4". 

solid  solid 

By  aid  of  analysis,  and  conductivity  and  electromotive 
force  observations.  Findlay  found  the  ratio  —  ^—   to  be  a 


346  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

constant  for  all  such  systems,  and  to  have  a  value  at 
25°  lying  between  0.25  and  0.3.  The  outcome  of  the 
investigation  may  be  summed  up  as  follows:  From  a 
mixed  solution  of  sodium  iodide  and  sodium  sulphate,  by 
the  addition  oj  a  soluble  lead  salt,  pure  lead  iodide  (the 
more  soluble]  can  be  precipitated  if  the  ratio  oj  the  square 
of  the  concentration  of  iodine  ions  to  the  concentration  of 
the  sulphate  ions  is  greater  than  the  equilibrium  constant. 
When  the  ratio  becomes  equal  to  this  constant,  both  lead 
iodide  and  sulphate  are  precipitated  together,  the  ratio 

c\, 


-  remaining  constant.    And  all  this   is   true  for  the 
" 

sulphate  when  the  ratio  is  smaller  than  the  constant. 

It  will  be  observed  from  these  examples  that  by  aid 
of  the  law  of  mass  action,  even  though  it  fails  to  hold 
for  strong  electrolytes,  we  can  forsee  and  regulate  many, 
if  not  most,  of  the  reactions  with  which  we  come  in  con- 
tact. Many  other  examples  could  be  cited  here  to 
illustrate  the  methods  of  application,  but  the  few  above 
will  suffice  to  bring  out  the  general  principles,  and  enable 
the  reader  to  follow  work  of  this  sort.  Interesting 
cases  of  ionic  equilibria  have  also  been  studied  by  von 
Ende  (Dissertation  Gottingen,  1899),  Morse  (Zeit.  f. 
phys.  Chem.,  41,  709,  1902),  Sherrill  (Zeit.  f.  phys. 
Chem.,  43,  705,  1903),  Sherrill  and  Skowronski  (J.  Am. 
Chem.  Soc.,  27,  30,  1905),  Noyes  and  Whitcomb  (J, 


CHEMICAL  CHANGE.  347 

Am.  Chem.  Soc.,  27,  747,  1905),  and  Abel  (Zeit.  f.  anorg. 
Chem.,  26,  377,  1901). 

78.  The  color  of  solutions. — The  color  of  a  solution 
depends  apparently  upon  the  condition  of  the  solute  in 
the  solvent.  If  a  substance  is  not  at  all  ionized,  or 
but  slightly  so,  any  color  it  may  possess  must  be  attrib- 
uted to  the  un-ionized  substance.  In  case  the  ionization 
is  practically  complete  the  color  of  the  solution  will  be 
the  result  of  the  mixture  of  the  colors  of  the  ions;  or  if 
only  one  is  colored  that  color  will  be  the  color  of  the 
liquid.  When  partly  dissociated,  then,  the  color  of  a 
solution  will  be  the  result  of  the  mixture  of  the  colors 
of  the  ions  and  the  un-ionized  portion;  or  if  only  one  of 
these  is  colored  that  color  will  be  the  color  of  the  liquid. 
The  un-ionized  portion  in  cases,  however,  may  also  show 
the  color  of  the  ion;  see  Noyes,  Technology  Quarterly, 
17,  306,  1904. 

There  is  always  a  chance  of  error  here  if  the  color 
of  the  solid  is  assumed  to  be  its  color  in  solution.  The 
color  of  a  crystal,  for  example,  is  very  often  different 
from  that  of  the  substance  in  the  form  of  powder,  and, 
further,  it  is  possible  that  a  dissociation  takes  place 
in  the  water  of  crystallization.  In  this  latter  case,  of 
course,  the  solid  would  exhibit  the  same  color  as  the 
colored  ion.  The  only  correct  way  to  find  the  color 
of  the  undissociated  portion  in  solution  is  to  use  a  solvent 
in  which  the  substance  is  not  dissociated  to  any  extent; 


348  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

then  the  color  can  be  directly  observed.  This  is  not  difficult 
to  carry  out,  for  all  solvents  have  a  different  dissociating 
power,  and  either  alcohol,  ether,  benzene,  chloroform, 
or  acetone  will  be  found  to  serve  the  purpose. 

The  ions  of  most  acids  are  colorless ;  consequently  all 
salts  of  a  metal  in  very  dilute  solutions  will  have  the 
same  color,  i.e.,  the  color  of  the  metallic  ion.  In  more 
concentrated  solutions  this  is  not  true,  for  many  un- 
ionized substances  are  colored  and,  as  they  are  now 
present  to  a  greater  amount,  the  color  of  the  solution 
is  the  result  of  the  mixture  of  these  and  the  ions.  An 
example  of  this  is  given  by  solutions  of  cuprous  chloride, 
where  the  color  of  the  un-ionized  portion  is  yellow. 
But  the  copper  ion  is  blue,  hence  the  color  of  a  solution 
of  cuprous  chloride  may  be  either  yellow,  green,  or 
blue,  according  as  it  is  undissociated  or  ionized  to  a 
lesser  or  greater  degree.  All  copper  solutions  when  very 
dilute,  provided  the  negative  ion  is  colorless,  show  the 
same  blue  color. 

The  formation  of  a  complex  ion  can  be  followed  very 
closely  when  it  is  composed  of  a  colored  and  a  colorless 
ion.  Thus  if  a  KCN  solution  is  added  to  a  colored 
copper  solution  the  color  instantaneously  disappears,  due 
to  the  formation  of  the  ion  CuCN".  The  formation  of  a 
complex  ion  can  be  proven  -in  this  way,  but  its  nature  can 
only  be  shown  by  migration  experiments,  as  was  men- 
tioned above.  (See  Chapter  IX.) 


CHEMICAL  CH/tNGE.  349 

79.  The  action  of  indicators. — An  indicator  is  a  sub- 
stance which  possesses  a  different  color  in  an  alkaline 
solution  from  what.it  does  in  acid.  The  indicators  are 
themselves  slightly  dissociated  acids  or  bases,  and  the 
change  in  color  is  due  simply  to  the  rise  or  disappear- 
ance of  the  colored  ion  or  un-ionized  substance.  In- 
stead of  attempting  to  find  a  general  rule  as  to  their 
behavior,  we  shall  consider  a  few  typical  cases  which 
will  make  the  principle  clear. 

Phenol- phthalein  is  a  very  weak  acid,  i.e.,  in  water  it  is 
ionized  to  an  imperceptible  extent  into  H'and  the  colored 
negative  radical.  In  the  un-ionized  state,  i.e.,  in  an 
alcoholic  solution,  it  is  colorless,  while  the  negative  ions 
are  red.  If  potassium  hydrate  is  added  to  a  colorless 
alcoholic  water  solution,  the  following  reaction  takes 
place:  The  indicator  is  but  slightly  dissociated  into  H' 
ions,  but  when  these  come  into  contact  with  ions  of 
OH'  they  unite  with  them  to  form  undissociated  water. 
This  is  due  to  the  fact  that  only  an  infinitesimally  small 
amount  of  H*  and  OH'  ions  can  exist  together  without 
forming  undissociated  water.  This  removal  of  H*  ions, 
however,  destroys  the  equilibrium  which  exists  between 
the  ions  and  the  undissociated  portion  of  the  indicator 
(by  equation  46);  consequently  more  of  the  indicator 
dissociates.  Again  the  H*  ions  are  removed,  etc.,  and  the 
process  is  repeated  until  we  have  only  ions  of  K*  and 
the  negative  colored  radical,  and  the  solution  is  red, 


35°  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

It  is  a  general  rule  that  all  weak  acids  are  very  much 
less  dissociated  than  their  sodium  or  potassium  salts. 
In  few  words,  then,  the  process  consists  in  the  removal 
of  the  H*  ions  by  those  of  OH',  and  the  formation  of 
the  potassium  salt  of  the  indicator,  which  is  so  largely 
dissociated  that  the  red  color  of  the  negative  ion  is  visible. 

If  acid  is  added  to  this  salt,  then  ions  of  H*  come 
in  contact  with  those  of  the  negative  radical,  and  since 
they  cannot  exist  with  them  to  the  extent  that  those  of 
K'  can,  they  unite  again  to  form  the  almost  undisso- 
ciated  indicator,  and  the  color  disappears. 

Ammonium  hydrate  is  a  very  weak  base,  i.e.,  its  dis- 
sociation is  very  small  (about  1.5%  in  a  n/io  solution), 
and  the  salt  formed  from  it  with  the  very  weak  acid  phenol- 
phthalein  is  hydrolytically  dissociated  to  such  an  extent 
that  it  is  colorless,  until  such  an  excess  of  ammonia  is 
added  that  the  hydrolytic  dissociation  is  no  longer  possible. 

When  titrated  with  phenol-phthalein,  all  acids  give 
very  satisfactory  results,  because  the  number  of  H*  ions 
in  the  indicator  is  very  small,  and  therefore  easily 
influenced.  In  general,  then,  with  phenol-phthalein  only 
the  stronger  alkalies  can  be  used,  but  both  strong  as 
well  as  weak  acids  may  be  readily  determined  (contrast 
with  methyl  orange). 

Methyl  orange  is  a  medium  strong  acid,  i.e.,  it  is  disso- 
ciated to  a  larger  extent  than  phenol-phthalein.  Its 
un-jonized  portion  is  red  and  its  negative  ion  yellow. 


CHEMICAL   CHANGE.  351 

The  pure  water  solution  of  this  is  so  much  dissociated 
that  it  shows  the  color  which  is  produced  by  the  mixture 
of  these  two  colors;  for  this  reason  it  is  usually  used 
in  an  alcoholic  solution.  Addition  of  a  strong  acid, 
by  the  action  of  its  H"  ions,  causes  the  dissociation  to 
decrease  and  consequently  the  red  color  of  the  un-ionized 
substance  appears.  If  an  alcoholic  solution  which  is  red 
is  mixed  with  a  base,  H'  and  OH'  ions  unite  to  form 
H20,  and  a  dissociated  salt  is  left,  which  shows  the 
yellow  color  of  the  negative  ion.  A  very  weak  acid 
will  have  no  effect  upon  this  indicator,  for  the  number 
of  its  H'  ions  will  not  be  great  enough  to  influence  those 
of  the  indicator  (for  example,  H^COs). 

Methyl  orange,  then,  is  adapted  to  all  baces,  weak 
as  well  as  strong,  but  only  to  strong  acids  (contrast  with 
phenol-phthalein) . 

The  other  acid  indicators  act  on  the  same  principle  as 
these.  The  alkaline  indicators  possess  OH'  ions  and 
form  salts  with  the  acids,  so  that  the  color  of  the  positive 
ion  appears.  With  bases  the  ionization  is  decreased, 
hence  the  color  of  the  un-ionized  product  is  shown. 

In  all  cases  it  is  to  be  remembered  that  a  certain  por- 
tion of  the  acid  or  base  used  is  employed  in  causing  the 
indicator  to  change,  so  that  the  amount  of  the  latter  should 
be  as  small  as  possible. 

The  action  of  acids  in  each  of  these  cases  is  to  decrease 
the  ionization  by  its  H*  ions,  or  else  to  remove  OH'  ions 


35*          ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

with  them.  The  bases  decrease  the  concentration  of 
OH'  ions,  or  remove  H*  ions. 

This  theory  is  due  to  Ostwald  and  apparently  repre- 
sents the  facts.  Another  theory  has  recently  been 
advanced  by  Stieglitz  (J.  Am.  Chem.  Soc.,  25, 1112,  1903), 
but  as  neither  can  be  proven  absolutely  ^  and  as  the  one 
above  accounts  for  the  facts  observed  more  simply  than 
the  other,  it  may  be  provisionally  accepted. 

80.  General  analytical  reactions. — Here  we  shall  not 
treat  the  analytical  scheme  in  detail,  but  only  consider 
those  reactions  which,  viewed  from  the  standpoint  of 
the  theory  of  dissociation,  possess  particular  interest. 

METALS    OF    THE   SECOND  GROUP. 

Calcium  oxalate  is  a  comparatively  insoluble  sub- 
stance ;  its  solubility  is  0.000059  mole  per  liter  at  25°, 
so  that  s  =  3.3Xio~9.  Oxalic  acid  is  not  dissociated 
enough,  however,  to  precipitate  CaC2O4  completely  from 
the  salts  of  strong  acids.  This  is  due  to  the  fact  that 
the  reaction  as  it  progresses  gives  rise  to  ions  of  H*  and 
of  the  negative  radical,  i.e.,  to  a  strong  acid,  and  the 
H'  ions  of  this  drive  back  the  dissociation  of  the  oxalic 
acid  to  such  an  extent  that  the  solubility  product  of 
CaC2O4  is  no  longer  exceeded,  even  with  the  large 
excess  of  Ca"  ions  present.  In  contrast  to  this,  all  cal- 
cium salts  are  precipitated  by  ammonium  oxalate,  since 
that  is  dissociated  to  a  larger  degree,  and  no  acid  is  formed 


CHEMICAL  CHANGE.  353 

to  affect  its  dissociation.  If,  however,  NaCl  is  added  to 
CaCU,  more  CaC2C>4  is  precipitated  by  H2C2O4,  since 
the  Cr  ions  of  the  NaCl  drive  back  the  dissociation  of 
the  HC1  formed  during  the  reaction,  thus  decreasing  the 
influence  upon  the  dissociation  of  the  oxalic  acid. 

Ammonia  is  too  weak  a  base  (1.5%  ionized  in  a  n/io 
solution)  to  form  the  hydrate  of  calcium,  since  the  number 
of  its  OH'  ions  is  too  small  to  cause  the  solubility  pro- 
duct of  the  Ca(OH)2  to  be  exceeded,  even  with  the 
excess  of  Ca"  ions.  The  hydrates  of  sodium  and  potas- 
sium, however,  are  dissociated  to  a  greater  degree,  and 
the  precipitate  is  formed.  Ca(OH)2  is  soluble  in  water 
to  the  extent  0.035  mole  per  liter,  but  an  excess  of  the 
precipitant  will  reduce  this  so  that  Ca(OH)2  may  be 
used  as  a  means  to  detect  calcium.  Of  course  the  OH' 
ions  here  have  a  greater  effect  upon  the  Ca"  than  those 
of  a  divalent  element  would  have,  for  here  in  determining 
the  solubility  product  the  square  of  the  concentration 
of  OH'  ions  is  used.  We  have  thus  in  a  saturated  so- 
lution, since  Ca"  =  0.035  and  0^  =  2X0.035  =  0.07, 
s  =  0.035  X  (0.07)2  =  0.031 72.  Thus  i  mole  of  OH'  ions 
reduces  the  solubility  of  Ca(OH)2  to  0.031 72  =  .r(#+i)2, 
or  since  the  x  in  x+i  may  be  neglected  as  compared 
with  i,  #(i)2  =  o.03i72,  i.e.,  the  solubility  is  reduced  by  a 
normal  solution  of  OH'  ions  from  0.035  to  0.03172.  Con- 
sidering that  the  compound  CaY  has  the  same  solu- 
bility as  Ca(OH)2  and  that  Y  is  divalent,  we  find  0.035 


354  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

X 0.035  =  51  =  0.02123.  In  this  case  a  normal  solu- 
tion of  Y  ions  would  depress  the  solubility  from  0.035 
only  to  x  =  0.02!  23. 

Strontium  salts  may  be  precipitated  as  the  sulphate, 
but  as  this  is  quite  soluble  it  is  necessary  to  add  alcohol 
and  an  excess  of  the  precipitant  to  decrease  the  disso- 
ciation and  consequently  the  solubility.  Since  H2SO4 
is  ionized  to  a  lesser  degree  than  HC1  and  HNO3,  the 
sulphate  is  soluble  to  a  certain  extent  in  them.  The 
process  here  is  as  follows:  The  SrSC>4  gives  off  ions  of 
Sr"  and  SO/';  these  latter,  however,  owing  to  the  large 
concentration  of  H'  ions  in  the  acid,  unite  to  form 
HSCV  or  H2SO4-  This  loss  of  SCV'  ions  causes  more 
SrSC>4  to  dissolve  and  dissociate,  so  that  the  SrSC>4  dis- 
solves more  than  it  does  in  pure  water. 

If  a  large  excess  of  a  solution  of  a  soluble  carbonate 
is  poured  upon  dry  SrSC>4  the  latter  is  transformed 
into  SrCO3.  This  is  due  to  the  fact  that  SrCO3  has  a 
much  smaller  solubility  product  than  SrSC>4.  In  the 
solution  we  have  ions  of  Sr",  SCV,  2Na*,  and  CO3";  the 
Sr*  ions  unite  with  those  of  CO  3"  to  form  SrCO3,  which 
saturates  the  solution  and  then  separates  out  as  a  solid. 
This  causes  more  SrSC>4  to  dissolve  and  dissociate,  which 
in  its  turn  is  transformed  into  the  carbonate,  until  finally 
we  have  solid  SrCO3  in  a  solution  of  NaSC>4. 

The  solubility  product  of  SrCO3  is  so  much  smaller 
than  that  of  the  SrSC>4  that  SrSC>4  is  transformed  com- 


CHEMICAL   CHANGE.  355 

pktely  by  a  mixture  of  equal  amounts  of  carbonate 
and  sulphate  for  the  decrease  of  the  dissociation  of  the 
SrSO4  by  the  SO4"  ions  is  not  sufficient  to  prevent  the 
formation  of  the  SrCO3.  If  the  solubility  product  of  a 
sulphate  differs  less  from  that  of  the  carbonate,  as  in 
the  case  of  barium,  then  the  sulphate  when  treated  with 
a  mixture  of  equal  amounts  of  a  carbonate  and  sulphate 
will  not  be  changed.  This  is  true  because  the  SO  4"  ions 
added  decrease  the  dissociation  to  such  an  extent  that 
the  solubility  product  of  the  carbonate  is  not  reached. 
We  have  so  much  ionized  SO/'  in  the  solution  that  the 
metal  ions  with  those  of  the  CO3"  could  only  reach  the 
solubility  product  of  the  carbonate  if  it  were  considerably 
smaller  than  that  of  the  sulphate.  This  is  a  method  of 
separating  Sr  from  Ba,  for  the  SrCO3  formed  is  soluble 
in  HC1,  while  the  unchanged  BaSO4  is  not. 

Morgan  (J.  Am.  Chem.  Soc.,  21,  522, 1899)  has  shown 
this  relation  to  be  as  follows   (the  solubilities  refer  to 

18°): 

Ba**XCO3"  =  5i  =  o.o3iXo.o3i=o.o7i 
Ba"X  SO4"  =  S2  =  0.041X0.041  =  0.091 
Sr"  X  CO3"  =  s3  =  o.o47.Xo.o47  =  o.o85 
Sr"  X  SO4"  =  s4  =  0.0554  X o.o354  =  o.o63 

To  precipitate  BaCO3  from  i  liter  of  BaSO4  solution 
we  must  have  more  than 

0.041  Xx = 0.071,     x= 0.02!  of  CO3"  ions. 


35<5  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

For  SrCOa  from  SrSO4  more  than 
0.0354  X#=o.o85, 

#=0.0592  of  COa"  ions. 

If  we  add  SO/'  ions  to  the  concentration  of  o.i  mole 
to  a  BaSC>4  solution  then  the  amount  of  BaSC>4  remain- 

ing will  be 

;yX.i=o.o9i, 

y  =  o.osi=  moles  of  Ba",  of  SCV,  and  of  course  of  BaSC>4, 
in  solution,  i.e.,  the  new  solubility  of  BaSC>4. 

If  we  now  find  the  concentration  of  COa"  ions  neces- 
sary to  reach  the  solubility  product  of  BaCOa  we  find 
it  is  increased  to 

o.o8iX#=o.o7i, 

oc=io  moles  from  0.^21  moles, 

i.e.,  in  the  case  of  Ba  a  o.i  molar  solution  of  SCV  ions 
counteracts  the  effect  of  10  moles  of.  COa"  ions.  For 
Sr'we  find  this  to  be  in  the  two  cases 

yX.  i  =0.063, 

:v=  0.053, 

i.e.,  the  solubility  of  SrSCU  in  presence  of  .1  SCV  ions. 
From  this  in  order  to  precipitate  SrCOa  it  is  necessary 
to  have  only 


moles. 

That  is,  the  effect  of  SCV  ions  on  the  formation  of 
SrCOa  from  SrSC>4  is  so  small  that  Sr  will  be  trans- 


CHEMICAL  CHANGE.  357 

formed  by  solutions  of  COs"  ions  which  have  no  effect 
upon  BaSO4;  and  the  NaSO4  formed  during  the  reaction 
increases  this  difference  between  Sr  and  Ba. 

In  the  case  of  the  carbonates  here  the  values  are  very 
largely  affected  by  the  hydrolytic  dissociation.  This, 
however,  would  have  the  same  effect  upon  both  Sr  and 
Ba,  so  the  relation  between  the  two  holds  just. as  well 
as  if  this  did  not  take  place.  Bodlander  has  lately 
determined  the  hydrolytic  dissociation  of  BaCO3  and 
CaCOs  and  finds  them  equally  dissociated,  i.e.,  to  84%. 

Strontium  sulphate  is  characterized  by  its  extreme 
slowness  of  saturation,  so  that  BaSO4,  which  is  formed 
immediately,  can  be  separated  from  the  SrSO4,  which 
is  formed  only  after  a  time. 

As  will  be  observed  this  process  is  similar  in  principle 
to  the  later  work  of  Findlay  described  on  page  346. 

Magnesium. — The  hydrate  of  this  metal  is  slightly 
soluble  (0.00002  moles  per  liter),  but  an  excess  of  OH' 
ions  reduces  the  solubility  very  largely,  so  that  it  may 
be  used  a  quantitative  precipitate.  The  fact  that  Mg" 
and  OH'  ions  can  exist  to  an  extent  together  without 
uniting  explains  why  Mg(OH)2  is  not  precipitated  by 
NH4OH,  when  it  is  by  NaOH  and  KOH.  The  NH4OH 
contains  just  enough  OHr  ions  to  -exceed  the  value  of 
the  solubility  product  of  Mg(OH)2,  but  the  ammonium 
salt  which  is  formed  by  the  reaction  has  a  decreasing 
effect  upon  the  dissociation  of  the  NH4OH,  so  that 


358  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

the  product  can  no  longer  be  exceeded.  In  this  way 
an  ammonium  salt  added  to  one  of  magnesium  will 
prevent  the  precipitation  of  the  latter  as  Mg(OH)2  by 
NH4OH,  and  also  by  other  hydrates,  unless  enough  is 
first  added  to  saturate  all  the  NH4*  ions  in  the  solution. 
Undissociated  NH4OH  will  be  formed  from  the  NH4' 
and  OH'  ions  and  an  excess  of  OH'  ions  can  never  be 
present  until  all  NH4'  ions,  except  those  few  produced 
from  the  NH4OH,  are  removed.  For  the  proof  of  this 
see  above,  pages  339~344- 

METALS  OF  THE  THIRD  AND  FOURTH  GROUPS. 

These  groups  of  metals  are  characterized  by  the  fact 
that  their  sulphides  have  such  large  solubility  products 
that  they  are  soluble  in  dilute  acids.  A  general  law 
as  to  the  solubility  of  the  sulphides  in  acids  is  as  follows, 
its  derivation  being  self-evident:  //  the  solution  of  H2S 
gas  in  acid,  such  as  would  be  formed  if  the  sulphide  dis- 
solved in  acid  of  a  certain  strength,  contains  a  smaller 
number  of  5"  ions  than  there  are  in  a  water  solution  oj 
the  sulphide,  then  the  sulphide  is  soluble  in  acid.  If  a 
larger  number,  then  the  sulphide  is  insoluble.  If  the  sul- 
phide contains  in  H2O  a  larger  number  of  S"  ions  than 
H2S  in  an  acid  solution,  then  when  it  is  placed  in  acid  the 
solubility  product  of  the  H2S  is  overstepped  and  H* 
and  S"  ions  unite,  saturate  the  solution,  and  finally  escape 
as  H2S  gas.  If  the  sulphide  contains  a  smaller  number 


CHEMICAL   CHANGE.  .        359 

of  S"  ions  in  water,  then  it  will  be  impossible  for  the 
solubility  product  to  be  overstepped  and  no  H2S  gas  will 
be  formed. 

Iron. — This  metal  has  two  ionic  forms — one  with 
two  charges  of  electricity,  Fe",  and  the  other  with  three, 
Fe"".  The  solubility  product  of  FeS  is  so  large  that 
it  is  not  formed  by  H2S  in  neutral  salt  solutions,  for  the 
acid  formed  drives  back  the  dissociation  of  the  H2S 
(which  is  ionized  to  a  very  small  extent  into  2H*  and 
S"),  so  that  the  S"  ions  become  too  small  in  concentration 
to  exceed  the  solubility  product  of  the  FeS. 

The  iron  salts  of  the  strong  acids  are  hydrolytically 
dissociated;  in  the  case  of  salts  of  weak  acids  this  is 
almost  complete.  On  account  of  this  we  have  difficulty 
in  obtaining  clear  solutions  except  in  the  presence  of  acids, 
for  the  hydrate  formed,  FeOH",  or  whatever  formula 
it  may  possess,  is  in  the  colloidal  state  and  separates 
out  readily.  The  so-called  basic-acetate  separation  of 
iron  is  based  upon  this  dissociation.  The  iron  salt 
is  diluted  and  nearly  neutralized  with  a  carbonate,  and 
then  acetic  acid  and  an  acetate  added,  and  the  whole 
solution  boiled  and  filtered  hot.  This  treatment  causes 
the  acetate  of  iron  to  be  formed,  and  when  hot,  since 
the  hydrolytic  dissociation  increases  writh  the  temperature, 
the  hydrate  of  iron  separates  out  completely.  The  ace- 
tate is  added  of  course  to  depress  the  dissociation  of  the 
small  amount  of  acetic  acid  present  and  thus  to  destroy 


360       .      ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

its  dissolving  power.  In  this  way  it  is  possible  to  pre- 
cipitate iron  as  a  hydrate  without  making  the  solution 
alkaline. 

This  hydrolytic  dissociation,  of  course,  can  always  be 
predicted  by  the  use  of  the  equation  of  the  solubility 
product,  as  has  already  been  shown  on  pages  316-329. 

The  ferro  and  ferri  ions  have  different  colors,  the 
former  being  pale  greenish,  the  latter  pale  cream.  The 
brown  color  is  due  to  the  colloidal  hydroxide.  H2S  gas 
reduces  the  ferri  to  ferro  ions,  with  the  liberation  of 
sulphur.  By  boiling  iron  salts  with  KCN,  the  ferro-  and 
ferricyanides  are  formed,  each  of  which  dissociates  into  a 
complex  ion,  so  that  neither  gives  a  reaction  for  iron. 

The  volumetric  method  of  estimation  of  iron  by  per- 
manganate of  potash  depends  upon  the  change  from 
ferro  to  ferri  ions.  Thus 

2KMnO4  +  ioFeSO4  +  8H2SO4  =  K2SO4 

+  2MnSO4  +  5Fe2(SO4)3  +  8H2O, 

or,  since  it  is  an  ionic  reaction, 

ioFe"  +  2MnO4'  +  i6H'  =  ioFe'"  +  2Mn"  +  8H2O. 


The  end  of  the  reaction  is  the  point  at  which  the  color 
of  the  undecomposed  KMnO4  is  observed. 

Aluminium  has  only  trivalent  ions,  Al"'.  All  salts 
react  acid,  showing  hydrolytic  dissociation,  which  with 
the  weaker  acids  is  nearly  complete,  as  in  the  case  for  iron. 


CHEMICAL   CHANGE,  361 

The  behavior  of  A1(OH)3  with  ammonia,  caustic  soda, 
and  potash  is  just  the  opposite  to  that  of  Mg(OH)2. 
The  two  latter  dissolve  the  hydrate,  while  the  former 
does  not.  This  is  due  to  the  fact  that  with  the  stronger 
alkalies,  i.e.,  those  which  contain  a  large  concentration 
of  OH'  ions,  the  AT"  ion  takes  the  place  of  the  hydrogen 
in  the  OH',  and  we  have  the  complex  ion  AlOa"'.  This 
causes  more  of  the  A1(OH)3  to  dissolve  and  dissociate, 
etc.  With  ammonia  the  concentration  of  OH'  ions  is 
too  small  to  cause  the  ion  to  be  formed  to  any  extent; 
hence  the  A1(OH)3  is  not  so  soluble. 

Cobalt  and  nickel. — The  sulphides  of  these  two  metals 
show  a  marked  peculiarity.  They  are  not  formed  in 
dilute  acid  solutions,  and  yet  when  once  formed  are 
not  dissolved  by  acid  of  this  strength.  This  is  probably 
to  be  explained  by  the  fact  that  the  sulphides  when  once 
formed  change  their  state  in  some  way  and  thus  become 
insoluble  in  acids.  This  seems  to  be  a  common  process 
with  the  sulphides,  cadmium  sulphides  being  the  only 
one  which  is  perfectly  reversible.  This  is  formed  in 
a  solution  and  then  redissolved  by  removing  the  H2S 
gas  in  a  vacuum — a  process,  which  is  not  possible  experi- 
mentally with  any  other  metal,  although  theoretically 
possible  with  all.  With  all  metals  of  these  groups  H2S 
under  higher  pressure  will  cause  the  sulphides  to  be 
formed  in  the  presence  of  acid  which  would  prevent  the 
formation  under  atmospheric  pressure. 


362  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

Zinc  forms  with  the  alkalies  a  complex  ion,  ZnO2", 
just  as  aluminium.  The  sulphide  of  this  metal  is  as 
insoluble  as  any  of  this  group,  for  neutral  salt  solutions 
are  practically  (i.e.,  90  to  95%)  precipitated  by  H2S, 
notwithstanding  the  H*  ions  of  the  acid  formed  during 
the  reaction.  The  acetate  in  presence  of  a  neutral  acetate 
is  completely  precipitated  by  H2S. 

METALS   OF  THE  FIFTH   AND   SIXTH  GROUPS. 

The  sulphides  of  the  metals  of  these  groups  possess  a 
concentration  of  S"  ions  in  a  saturated  water  solution 
which  is  smaller  than  that  of  H2S  under  the  same  con- 
dition, even  in  the  presence  of  dilute  acids.  Conse- 
quently they  are  precipitated  in  acid  solutions  by  H2S  gas. 

Cadmium.  —  The  halogen  compounds  of  this  metal  are 
characterized  by  their  slight  ionization.  The  effect  of 
this  upon  the  solubility  is  marked,  for  when  no  dissocia- 
tion takes  place  the  precipitate  is  usually  soluble  in  an 
excess  of  the  precipitant.  As  a  rule,  complex  ions  con- 
taining Cd  are  not  stable,  i.e.,  their  dissociation  is  com- 
parable with  that  of  the  simple  salts.  The  sulphide  of 
cadmium  is  less  soluble  than  that  of  zinc  and  is  formed 
from  the  complex  salt  K2CdCN4  by  H2S  gas. 

Copper  can  be  estimated  from  the  cuprous  iodide. 
The  reaction  is 


In  order  to  make  the  Cul  as  insoluble  as  possible  it  is 


CHEMICAL   CHANGE.  363 

necessary  to  have  ions  of  I'  present.  An  addition  of 
H2SO3  serves  this  purpose,  in  that  it  forms  HI,  from 
the  undissociated  I,  which  is  dissociated  into  H*  and 
I'  ions. 

Copper  sulphide  is  somewhat  more  soluble  than  cad- 
mium sulphide,  and  too  much  acid  will  entirely  prevent 
the  precipitation.  An  excess  of  water,  however,  causes 
it  to  form. 

Mercury  halides  are  not  largely  ionized,  and  conse- 
quently many  of  the  reactions  take  place  between  the 
un-ionized  aggregates.  The  cyanide  of  mercury  does 
not  conduct  the  electric  current  in  solution,  i.e.,  is  ionized 
only  to  a  very  slight  extent,  at  most. 

As  hydrolysis  take  place  to  a  very  large  extent  with 
the  mercury  salts  of  the  oxygen  acids  (i.e.,  H2SO4, 
HNOs,  etc.),  an  addition  of  free  acid  is  necessary  if  a 
clear  solution  is  to  be  obtained. 

Bismuth  is  hydrolytically  dissociated  in  its  solutions 
to  a  greater  degree  than  any  other  metal,  and,  indeed, 
this  property  is  taken  advantage  of  in  its  separation  from 
other  substances.  If  a  salt  solution  in  acid  is  diluted 
with  water  it  is  noticed  that  a  basic  salt,  probably  BiOCl, 
is  immediately  precipitated. 

For  further  details  on  this  subject  the  reader  is  referred 
to  Ostwald's  Foundations  of  Analytical  Chemistry  or 
Bottger's  Qualitative  Analyse. 


CHAPTER  IX. 
ELECTROCHEMISTRY . 

A.  THE  MIGRATION  OF  THE  IONS. 

8 1.  Electrical  units. — The  unit  of  the  resistance 
offered  to  the  electric  current  is  the  ohm,  i.e.,  the  resist- 
ance at  a  temperature  of  o°  C.  of  a  column  of  mercury 
106.3  cms-  l°ng>  ^h  a  cross-section  equal  to  i  sq.  mm. 
This  is  equal  to  io9  absolute  units. 

The  unit  of  current  strength  is  that  strength  which 
will  separate  0.001118  gram  of  silver  from  a  solution  in 
one  second,  and  is  called  the  ampere — equal  to  lo"1 
absolute  units. 

The  unit  of  electromotive  force  is  the  volt,  or  io8 
absolute  units,  which  gives  a  value  for  the  Daniell  cell 
equal  to  i.io  volts. 

The  unit  of  the  amount  of  electricity  is  the  coulomb, 
i.e.,  amperes  per  second;  i  gram  of  H*  ions  carries  with 
it  96,537  coulombs.  This  number  is  called  the  Faraday, 

and  is  designated  by  F. 

364 


ELECTROCHEMISTRY.  365 

The  intensity  factor  of  electrical  energy  is  the  volt, 
while  the  capacity  factor  is  the  coulomb,  i.e., 


where  s  is  the  amount  of  current  in  coulombs,  TT  is  the 
electromotive  force  in  volts,  and  E  is  the  electrical  work 
the  unit  of  which  is  the  watt-second,  equal  to  the  volt- 
ampere-second,  i.e.,  equal  to  io7  absolute  units,  i  watt- 
second  is  the  electrical  work  done  when  i  ampere  flows 
at  the  potential  i  volt  for  i  second. 

The  heat  equivalent  of  electrical  energy,  since  the 
latter  is  equal  to  i  voltXi  coulomb  or  io7  absolute 
units,  is 


i.e.,  i  watt-second  =  0.2394  cal. 

To  separate  i  gram  of  H*  ions,  or  the  equivalent 
weight  in  grams  of  any  other  element  we  need  then  the 
amount  of  work 

rX  96540  X  watt-seconds  =  96  540X0.2394  XT: 
=  231107:  cals., 

where  n  is  the  electromotive  force  of  the  current  giving 
96,540  amperes  per  second. 

If  all  electrical  energy  is  transformed  into  heat,  then 


366  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

where  A  is  the  amount  of  heat  and  k  is  the  heat  equivalent 
of  electrical  work. 

82.  Faraday's  law. — This  is  the  basis  of  all  our  work  in 
electrochemistry.     The  law  may  be  expressed  as  follows: 

1.  The  amount  oj  any  substance  deposited  by  the  current 
is  proportional  to  the  quantity  oj  electrictiy  flowing  through 
the  electrolyte. 

2.  The  amounts  of  different  substances  deposited  by  the 
same  quantity  of  electricity  are  proportional  to  their  chemi- 
cal equivalent  weights. 

One  gram  equivalent  of  H*  ions  carries  with  it  96,540 
coulombs  of  electricity,  as  has  ..been  determined  by  ex- 
periment. One  coulomb,  then,  will  cause  0.041636 
gram  of  hydrogen  ions  to  separate;  hence  it  will  cause 
the  separation  of  0.041036  X  a  grams  of  any  other  element, 
where  a  is  the  equivalent  weight  of  the  element. 

83.  The  migration  of  the  ions. — The  chemical  effect  of 
the  passage  of  an  electric  current  through  an  electrolyte 
can  be  divided  into  two  distinct  portions,  viz.,  the   con- 
duction through  the  electrolyte,   and   the  separation  of 
substance   at   the   electrodes.      It  is    not  necessary  that 
the  ions  which  serve  for  the  conduction  of  the  current 
through  the  liquid  be  those  which  are  also  separated  at 
the   electrode,    for  secondary   reactions   may   take   place 
there,  causing  others  to  appear  as  the  result  of  the  electro- 
lysis.   It  is  to  be  remembered,  however,  that  even  in  such 
a   case   Faraday's   law  still   holds,  and   the  substances 


ELECTROCHEMISTRY  367 

separated  are  chemically  equivalent  in  amount  to  those 
which  would  have  been  separated  in  case  the  secondary 
action  had  been  avoided. 

Although  the  two  different  effects  are  observed  to- 
gether in  practice,  we  shall  consider  them  separately; 
taking  up  the  question  of  conduction  here,  and  putting  off 
that  of  separation  to  a  later  period. 

The  conduction  through  the  liquid  depends  upon  what 
we  have  designated  thus  far  as  ionized  matter,  and  varies 
according  to  the  mobility  of  this,  which,  in  turn,  is  de- 
pendent upon  the  specific  nature  of  the  matter,  its  con- 
centration, the  temperature,  and  the  nature  of  the  solvent. 

After  the  electrolysis  of  a  solution,  excluding,  or  allow- 
ing jor,  any  secondary  reaction,  it  is  found  experimentally 
that  the  concentrations  around  the  anode  and  cathode 
are  not  always  identical,  as  they  were  initially.  In 
some  few  cases  they  are,  it  is  true,  but  in  these  it  can  be 
shown  (as  will  be  done  later)  that  the  mobility,  i.e., 
velocity  through  the  liquid  at  a  certain  voltage,  is  the 
same  for  both  the  anion,  which  goes  to  the  anode,,  as  it 
is  for  the  cathion,  which  goes  to  the  cathode.  In  all 
other  cases  the  mobility,  or  speed  of  migration  through 
the  liquid,  is  different  for  the  two  kinds  of  ionized  matter 
of  which  the  substance  is  composed. 

And  further  than  this,  the  analysis  of  the  anode  and 
cathode  liquids  after  electrolysis,  excluding  secondary 
reactions,  leads  to  an  expression  for  the  relative  mobilities, 


368 


ELEMENTS  OF  PHYSICAL  CHEMISTRY. 


i.e.,  the  migration  ratio  of  the  two  ions.  Indeed,  when 
the  original  solution  is  colored,  this  difference  in  con- 
centration can  be  observed  qualitatively  by  the  eye.  The 
reasons  for  this  change,  together  with'  the  principle 
upon  which  its  quantitative  calculation  is  based,  will 
be  made  clear  by  the  following  considerations: 

Assume  the  vessel  in  Fig.  15  to  be  divided  into  three 
portions,  AC,  CD,  and  DB,  and  filled  with  a  solution 

A       C      D      B 


FIG.  15. 

containing  30  gram  equivalents  of  HC1.  We  have,  then, 
10  gram  equivalents  in  each  division.  If  96,540  cou- 
lombs of  electricity  are  passed  through  the  cell  from 
A  to  B,  i  gram  equivalent  of  H'  ions  and  i  gram  equiv- 
alent of  Cl'  ions  will  be  separated  upon  the  electrodes 
B  and  A,  if  secondary  action  is  excluded.  These 
gases  we  assume  to  be  removed  as  they  are  formed. 
These  96,540  coulombs  passing,  as  they  do,  through  the 
whole  solution  have  a  certain  effect  upon  the  equilibrium 
of  the  ions.  First  we  will  imagine  the  H'  and  Cl'  ions 
to  move  with  the  same  velocity  and  then  with  differing 
velocities,  and  find  the  relation  between  the  change  in 
concentration  and  the  relative 


ELECTROCHEMISTRY.  369 

It  is  to  be  remembered  here  that  two  oppositely 
electrified  bodies  composing  a  system  will  transport  a 
current  equal  to  the  sum  of  the  charges  carried  by  the 
two  bodies  in  the  opposite  direction,  for  a  negative 
charge  going  in  one  direction  is  equivalent  to  an  equal 
and  opposite  charge  going  in  the  contrary  direction.  In 
other  words,  all  of  the  current  may  be  transported  by 
the  positive  material,  or  a  portion  may  be  carried  by 
each  in  opposite  directions,  and  in  all  cases  the  total 
current  is  the  sum  of  those  currents  going  in  the  opposite 
directions. 

I.  If  the  velocity  for  each  ion  is  the  same,  then  1/2  gram 
equivalent  of  Cl'  ions,  charged  with  48,270  coulombs, 
will  migrate  from  BD  through  DC  to  CA]  and  1/2  gram 
equivalent  of  H*  ions,  with  the  same  amount  of  electricity, 
will  go  from  AC  through  CD  to  DB.  Altogether,  then, 
i  gram  equivalent,  charged  with  96,540  coulombs,  has 
passed  through  the  section  CD.  Since  i  gram  equiva- 
lent of  H*  ions  has  been  removed  by  decomposition  from 
BD  and  1/2  gram  equivalent  has  migrated  to  it,  we 
have  left  9}  gram  equivalents  of  H*  ions  and  9!  gram 
equivalents  of  Cl'  ions,  since  but  1/2  gram  equivalent 
of  this  has  migrated  from  it.  Consequently  we  have 
in  BD  9^  gram  equivalents  of  HC1.  In  AC  we  have 
the  same  number,  since  i  gram  equivalent  of  Cl'  ions 
has  disappeared  and  1/2  gram  equivalent  has  migrated 
to  it  and  1/2  gram  equivalent  of  H'  ions  has  migrated  from 


37°  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

it.  In  the  section  DC  the  concentration  is  unaltered, 
i.e.,  just  as  much  ionized  matter  has  left  it  as  has  been 
carried  to  it. 

The  concentration  at  the  anode  is  the  same  as  that  at 
the  cathode,  then,  ajter  the  electrolysis  of  a  solution  com- 
posed o)  ions  with  the  same  mobility. 

II.  Assume  the  velocity  of  the  H'  ions  to  be  five  times 
that  of  those  of  Cl'. 

In  this  case,  after  i  gram  equivalent  of  H  and  i  gram 
equivalent  of  Cl  have  separated  in  the  gaseous  state, 
the  whole  system  will  have  suffered  a  change.  5/6  of 
a  gram  equivalent  of  H"  ions,  charged  with  1(96,540) 
coulombs,  will  migrate  from  AC  through  DC  to  BD, 
and  1/6  of  a  gram  equivalent  of  Cl'  ions,  with  J  (96, 540) 
coulombs,  will  go  from  BD  through  CD  to  AC.  Al- 
together, as  before,  i  gram  equivalent  of  ions  will  go 
through  the  section  CD,  carrying  with  it  96,540  coulombs 
of  electricity. 

The  original  composition  of  the  solution  in  CD  is 
again  unchanged.  In  BD  we  have  lost  i  gram  equivalent 
of  H*  ions  in  the  form  of  gas,  and  gained  5/6  of  a  gram 
equivalent  by  the  migration;  consequently  we  have 
9^  gram  equivalents  of  H'  ions  left.  1/6  of  a  gram 
equivalent  of  Cl'  ions  has  migrated  from  it,  so  that  in 
DC  we  have  9!  gram  equivalents  of  HC1. 

In  A  C  we  have  lost  5/6  of  a  gram  equivalent  of  H* 
'ons  and  i  gram  equivalent  of  Cl'  ions  as  gas,  but  have 


ELECTROCHEMISTRY.  371 

gained  1/6  of  a  gram  equivalent  of  Cl'  by  the  migra- 
tion; consequently  we  have  9^  gram  equivalents  of 
HC1  left. 

From  these  two  examples  the  following  law  may  be 
deduced:  The  loss  on  the  cathode  (BD)  is  related  to 
that  on  the  anode  (AC)  as  the  mobility  0}  the  anion  (Cl') 
is  to  that  of  the  cathion  (H'). 

In  this  way  Hittorf  determined  the  relative  mobilities 
or  migration  ratios  of  the  various  ions. 

84.  Determination  of  the  migration  ratios. — The  prac- 
tical determination  of  the  relative  mobilities  is  merely 
a  matter  of  analysis.  The  apparatus  which  is  used 
for  this  purpose  is  a  decomposition-cell,  so  arranged 
that  no  metal  can  drop  from  one  electrode  to  the  other; 
or  a  U  tube  may  serve  the  purpose,  so  long  as  the  two 
portions  of  liquid  may  be  removed  and  analyzed  sepa- 
rately. The  apparatus  is  filled  with  solution  and  the 
current  passed  through  for  a  certain  length  of  time,  the 
electrodes  being  of  the  metal  which  is  contained  in  the 
salt  or  inert,  except  in  cases  where  certain  secondary 
actions  are  to  be  avoided.  After  a  certain  time  either 
the  anode  or  cathode  portion  is  withdrawn  and  analyzed. 
This  analysis  will  give  us  the  loss  of  metal  on  the  one 
electrode,  from  which  that  on  the  other  may  be  cal- 
culated. 

If  n  is  the  fraction  of  the  cathion  which  has  migrated 
from  the  anode  to  the  cathode  when  one  gram  equiva- 


372  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

lent  has  been  separated,  then  i—  n  is  that  fraction  of 
the  anion  which  has  gone  to  the  anode.  These  two 
quantities,  n  and  i—  n,  are  called  the  migration  ratios 
of  the  cathion  and  anion.  We  have  then 

n        loss  at  anode      Uc 


i—n    loss  at  cathode     Ud 

where  Uc  is  the  mobility  of  the  cathion  and  Ua  that  of 
the  anion. 

An  example  will  make  the  determination  of  this  clear: 
Hittorf  electrolyzed  a  solution  of  AgNO3  until  1.2591 
grams  of  Ag  were  separated.  The  volume  of  liquid  at 
the  cathode  before  the  experiment  gave  17.46249  grams 
of  AgCl,  and  after  it  but  16.6796  grams,  i.e.,  a  loss  of 
0.7828  gram  of  AgCl  or  of  0.5893  gram  of  Ag. 

If  no  Ag  had  come  to  the  cathode  by  migration,  the 
solution  would  have  lost  1.2591  grams  of  Ag;  it  lost, 
however,  only  0.5893  gram;  hence  1.2591—0.5893  = 
0.6698  gram  Ag  has  come  to  it  by  the  migration.  If 
just  as  much  of  the  Ag  had  come  by  migration  as  had 
been  separated,  the  migration  ratio  of  the  Ag  would 
have  been  i,  i.e.,  the  NOs  ions  would  not  have  migrated. 
Only  0.6698  gram  of  Ag  has  migrated,  however;  hence 
the  migration  ratio  for  the  Ag'  ions  in  AgNOs  can  be 
found  by  the  proportion 

1.2591 10.6698:  :i  ^=0.532; 


ELECTROCHEMISTR  Y.  373 

the  migration  ratio  of  the  NCV  ions  is 

1-0.532  =  0.468. 

These  values  are  not  the  same  for  all  dilutions,  al- 
though in  general  the  variation  is  but  slight. 

A  table  containing  a  large  number  of  results  from 
experiments  of  this  sort  is  given  by  Kohlrausch  and  Hol- 
born,  Leitvermogen  der  ElektrolytCj  a  few  of  which  will 
be  found  in  Table  XIV. 


TABLE  XIV. 

HITTORF'S  MIGRATION  RATIOS  FOR  THE  IONS. 
Solutions  i/io  equivalent  normal. 

Substance.  i  —  n 

NH4C1 0.508 

1/2  BaClj 0.61 

1/2  CaCla o .  68 

1/2  MgCl2 0.68 

HC1 0.21 

KNO, 


Substance.  i  -  n 

1/2  K£O4 0.60 

1/2  CuSO4 , o .  64 

I/2H2SO, 0.21 

i/2K2C03 0.37 

1/2  NajCOg 0.48 

i/2Li2C03 0.59 

KOH 0.74 

NaOH 0.84 

KC1 0.507 

NaCl 0.63 

LiCl   o .  70 


3 0.50 

NaN03 0.61 

AgNO3 0.526 

i/2Ca(NO3)2 0.61 

KCIO3 0.46 


Naturally,  inert  anodes  can  also  be  used  for  this  pur- 
pose, and  in  many  cases,  to  avoid  secondary  reactions, 
an  anode  which  is  different  from  the  metal  of  the  salt 
may  be  used.  Thus  Hittorf  (Ostwald's  Klassiker  der 
exakten  Wissenschaften,  Nos.  21  and  23)  used  cad- 
mium as  an  anode  in  determining  the  migration  ratio 


374  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  sodium  chloride,  to  avoid  the  H*  ions  due  to  the  reac- 
tion of  chlorine  on  water. 

When  the  substances  separated  at  the  electrodes  are 
known,  analysis  of  the  liquids  at  the  cathode  and  anode 
enable  us  to  find  the  nature  and  composition  of  the  ions 
into  which  the  substance  dissociates.  In  most  cases 
here  qualitative  observations  suffice  for  the  purpose,  and 
often  color-changes  make  it  possible  to  follow  the  process, 
even  without  analysis.  Examples  of  this  method  are 
to  be  found  in  many  investigations,  see,  for  example, 
Noyes  and  Blanchard  (J.  Am.  Chem.  Soc.,  22,  729  and 
732,  1900),  Peters  (Zeit.  f.  phys.  Chem.,  26,  229,  1898), 
Calvert  (ibid. ,  38,  535,  1901),  and  Morse  (ibid.,  41,  709, 
1902).  The  use  of  agar-agar  or  gelatine  in  these  cases 
enables  us  to  obtain  the  solution  in  the  form  of  a  jelly, 
its  conduction  relation  remaining  almost  unchanged 
(see  p.  194). 

B.  THE  CONDUCTIVITY  OF  ELECTROLYTES. 

85.  The  specific  conductivity.  —  In  measuring  the 
electrical  conductivity  of  a  solution  we  obtain  results 
in  two  different  units:  one  refers  to  the  same  amount 
of  all  solutions,  the  specific  conductivity;  the  other  refers 
to  the  equivalent  or  molecular  amount,  the  molecular, 
or  equivalent,  conductivity. 

The  unit  oj  conductivity  is  the  conductivity  which  a 
column  oj  a  length  of  i  cm.  and  a  cross-section  oj  i  cm.2 


ELEC  TROCHE  MIS  TRY.  375 

possesses  when  its  resistance  is  one  ohm.  The  best  con- 
ducting aqueous  solutions  of  strong  acids  have  this  con- 
ducjivity  at  about  40°.  The  conductivities  based  on 
this  unit  are  designated  by  /c. 

86.  Molecular  and  equivalent  conductivity.  —  Since  the 
conductivity  of  a  solution  depends  almost  exclusively 
upon  the  amount  of  substance  dissolved  in  it,  it  is  more 
convenient  for  us  to  express  our  results  in  molecular 
or  equivalent  terms. 

The  equivalent  (molecular)  conductivity  oj  a  substance 
is  the  conductivity  oj  the  solution  which  contains  i  equiva- 
lent (i  mole)  of  substance,  the  electrodes  being  separated 
by  i  cm.  and  large  enough  to  contain  between  them  the 
entire  solution.  This  value  can  be  found  by  dividing  K 
by  the  number  of  equivalents  (moles)  per  cubic  centi- 
meter, or  by  multiplying  «  by  the  number  of  cubic  centi- 
meters in  which  i  equivalent  (i  mole)  is  dissolved. 
When  V  is  the  volume  containing  i  equivalent  (or  i 
mole)  in  liters,  then 


and 


where  equivalent  conductivity  is  designated  by  A  and 
molecular  conductivity  by  //, 


376  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

87.  Determination  of  electrical  conductivity. — Since 
the  degree  of  dissociation  of  electrolytes  can  be  deter- 
mined most  accurately  by  aid  of  the  conductivity,  .the 
method  is  one  of  great  importance.  The  apparatus  used 
for  this  purpose  was  designed  by  Kohlrausch,*  and  is 
similar  to  that  for  determining  the  resistance  of  metals, 
except  that  an  alternating  current  is  used  in  place  of 
the  direct,  and  a  telephone  receiver  instead  of  the  gal- 
vanometer. The  alternating  current  is  used  here  to 
prevent  actual  decomposition  of  the  solution,  for  that 
would  decrease  its  concentration  continually  and  cause 
polarization  of  the  electrodes.  In  this  way  all  substance 
which  is  deposited  by  the  current  in  one  direction  is 
redissolved  by  it  in  the  opposite  direction,  and  all  polari- 
zation effect  is  nullified. 

The  form  of  the  electrode  is  shown  in  Fig.  16,  the 
connections  being  made  by  aid  of  the  mercury  in  the 
glass  tubes.  The  electrodes  themselves  are  of  Pt,  which 
are  coated  electrolytically  with  platinum-black,  so  as  to 
do  away  with  any  difference  of  potential  between  them. 

It  is  not  necessary  to  have  these  electrodes  just  a 
square  centimeter  in  cross-section  or  i  cm.  apart,  for 
we  can  easily  find  the  factor  which  will  transform  results 
for  any  cell  into  terms  of  specific  conductivity.  Kohl- 
rausch has  carefully  determined  the  specific  conductivity 

*  For  details  see  Ostwald,  Handbook  of  Physicochemical  Measure- 
ments, Macmillan  &  Co, 


ELECTROCHEMISTR  Y. 


377 


of  a  number  of  standard  solutions,  so  that  a  determina- 
tion of  any  one  of  these  in  a  larger  or  smaller  cell 
will  give  directly  the  factor  necessary  to  transform  re- 


FIG.  1 6.     (Natural  size.) 


suits  by  that  cell  into  actual  specific  conductivities, 
according  to  the  definition.  For  a  0.02  molar  solution 
of  KC1,  Kohlrausch  found  the  values  K, 30  =  0.002397, 
J, 8o=  119.85,  £350=0.002768  and  A2S°=  138.54. 


378  ELEMENTS  OF  PHYSICAL   CHEMISTRY, 

88.  Ionic  conductivities. — Since  the  ions  are  the  car- 
riers of  electricity  in  solution,  the  equivalent  conductivity 
at  any  dilution  divided  by  that  at  infinite  dilution,  i.e., 
when  the  substance  is  present  only  in  the  form  of  ions, 
will  give  us  the  degree  of  dissociation.  We  have  then 


Av 


This  conductivity  at  infinite  dilution  means  simply  that 
the  equivalent  conductivity  is  not  altered  by  further 
dilution.  This  maximum  value  for  the  equivalent  con- 
ductivity Kohlrausch  found  for  a  binary  electrolyte  to 
be  equal  to  the  sum  of  two  single  values,  one  of  which 
refers  to  the  anion  and  the  other  to  the  cathion.  This 
law  of  the  independent  migration  of  the  ions  shows  that 
conductivity  is  an  additive  property.  The  truth  of  this 
law  is  shown  in  Table  XV. 

TABLE  XV. 

MOLECULAR  CONDUCTIVITIES  AT  INFINITE  DILUTION. 

Ag 

109 

103 
83 

The  differences  of  two  corresponding  sets  of  numbers 
jn  the  vertical  rows,  and  of  any  two  in  the  horizontal  ones, 


K 

Na 

Li 

NH4 

H 

Cl  

.  ..  123 

103 

95 

122 

353 

NO3  

.  ..  118 

98 

35° 

OH  

228 

2OI 

C103  

-••  H5 

.  .  . 

C2H302  .  .  . 

•••  94 

73 

ELECTROCHEMISTRY.  379 

are  nearly  equal,  which  can  only  occur  when  the  result 
is  composed  of  two  single  and  independent  values. 

One  kind  of  ion,  then,  always  carries  the  same  amount 
of  electricity  with  its  own  velocity,  independent  of  the 
nature  of  its  companion  ion. 

The  equivalent  conductivity  at  infinite  dilution  is  con- 
sequently 


where  lc  and  la  are  the  equivalent  conductivities  of  the 
ions  of  the  substance  in  solution  at  infinite  dilution. 
We  have  then  (p.  372) 


lc  ,  la 

=  -r~     and     i -»  =  -:-, 


in  other  Words,    lc  =  nAx,    la=(i-n)A00. 

Thus  the  molecular  conductivity  at  infinite  dilution 
fiM  (equal  here  to  the  equivalent  conductivity  Aw) 
of  sodium  chloride  is  no,  while  ^  =  0.38  and  1—^  =  0.62, 
hence 

*•   '   X 

/c  =  o.38Xno=4i.8(Na-), 
la =0.62  Xi  10  =  68.2  (CIO, 

i.e.,  i   mole  of  Na'  ions  possesses  a  conductivity  of  41.8 
when  between  electrodes  i  cm.  apart  and  large  enough  to 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

contain  between  them  the  total  volume  of  solution  in  which 
the  sodium  ions  exist;  and  i  mole  of  Clr  ions  under 
the  same  conditions  has  a  conductivity  equal  to  68.2. 

In  all  solutions  in  the  same  solvent,  at  the  same  tem- 
perature, these  values  remain  constant,  so  that  it  is 
possible  for  us  to  calculate  what  the  conductivity  would 
be  for  any  substance  at  infinite  dilution.  This  is  of 
great  use  in  experimental  work,  for  it  is  npt  always 
possible  to  actually  reach  this  limiting  value  with  any 
degree  of  accuracy. 

A  table  of  such  results,  then,  enables  us  to  find  the 
limiting  value  of  the  conductivity  at  infinite  dilution, 
and  not  only  this,  for 


i.e.,  if  we  know  the  fraction  of  a  mole  0}  each  ion  present 
we  can  find  the  equivalent  conductivity  of  the  solution  at 
any  dilution  by  multiplying  the  sum  of  the  ionic  conduc- 
tivities by  the  degree  of  ionization. 

Since  most  neutral  salts  are  very  largely  ionized  the 
value  of  the  equivalent  conductivity  can  be  readily  de- 
termined. By  subtracting  the  value  for  the  metal  ion 
from  this  result  it  is  possible,  then,  to  find  the  ionic 
conductivity  of  the  acid  radical  which  is  present  as  the 
negative  ion.  In  the  table  below  are  given  a  few  ionic 
conductivities,  from  which  various  values  may  be  cal- 
culated. 


ELECTROCHEMISTR  Y. 


IONIC  CONDUCTIVITIES  AT  18°  AND  INFINITE 


381 


Li"  

/ 

•1-2   44 

Temp.  Coeff  . 
0.0265 

£  Zn"      .    .  . 

/ 
4?  6 

Temp.  Coeff. 
o  0251 

Na'  

AT.  .  re 

0.0244 

i  Mg" 

46  o 

o  02  c.6 

K" 

64  67 

o  02  1  7 

\  Ba  ' 

(-6     2 

Rb'  

67.6 

0.0214 

i  Pb".. 

61   c 

o  024^ 

Cs' 

68   2 

O    O2  12 

£  SO/' 

68  7 

NH/  
Tl- 

64.4 

66 

0.0222 

o  0215 

*C03"  
BrO'  .  . 

70 

46    2 

0.0270 

Ag'  
F'  

54.02 
46  .  64 

0.0229 
0.0238 

CIO/  
IO/.  

64-7 

47    7 



Cl'  

6?  .44 

0.0216 

MnO/  

tr-j   4 

Br' 

67  61 

o  02  1  5 

CHO  ' 

46   7 

I' 

66  40 

o  021  3 

C  H,O  ' 

7C 

SCN' 

«;6  6? 

O    O2  1  1 

C,H.O' 

TT 

CIO,' 

ec   Q-t 

O   O2I  5 

C,H,O/ 

27    6 

I03V  .  .  . 
NO  '    . 

jj  -wo 
•         33-87 

61  78 

0.0234 

0.0205 

C.H.O,'.... 

CeH,,O,  . 

25-7 
24    3. 



Et- 
on' ' 

•     3i3 

174 

The  ionic  conductivity  of  elementary  ions  is  a  periodic 
function  of  the  atomic  weight  and  rises  with  it  in  each 
series  of  analogous  elements;  but  analogous  elements  with 
atomic  weights  greater  than  35  have  approximately  equal 
conductivities. 

Isomeric  and  metameric  anions  have  the  same  equiva- 
lent conductivities. 

89.  The  conductivity  of  organic  acids. — The  deter- 
mination of  the  ionization  constant  of  organic  acid, 
which  has  been  discussed  above,  is  very  easily  accom- 
plished by  aid  of  the  conductivity.  By  substituting 

—  for  a  in  the  Ostwald  dilution  law  we  obtain 


=  KV. 


*  See  Kohlrausch,  Berl.  Akademieber.  26,  581,  1902,  and  Sitzungsber. 
d.  Akad.  d.  Wiss.  zu  Berlin,  26,  572,  1902, 


382  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

For  inorganic  acids,  bases,  and  salts,  for  which  the  Ost- 
wald  dilution  law  does  not  hold,  we  can  use  either  the 
Rudolphi  or  the  van't  Hoff  dilution  law,  which  become, 

by  substitution  of  —  for  a, 


=  const. 


-i) 


VF 

V     *J 
and 


(-1) 


2 

F 


=  const. 


The  molecular  conductivity,  when  the  acid  is  very 
weak  and  dilute,  is  found  from  the  experimentally  de- 
termined value  of  /c,  after  subtracting  from  it  the  value 
*H2o  for  the  water  used,  for  these  two  terms,  under 
these  conditions,  become  of  the  same  order. 

90.  The  absolute  mobility  of  the  ions. — Thus  far  we 
have  considered  only  the  relative  mobility  of  the  ions,  or 
else  their  conductivities ;  now,  however,  we  shall  consider 
the  actual  velocity  in  centimeters  per  second  with  which 
they  go  through  the  solution. 


ELECTROCHEMISTRY.  383 

Imagine  two  electrodes  i  cm.  apart  having  a  difference 
of  potential  equal  to  i  volt,  and  assume  between  them 
a  solution  containing  i  equivalent  mole  of  positive  and 
i  equivalent  mole  of  negative  ions.  In  i  second  the 
amount  of  electricity  e  will  go  over.  Each  equivalent 
mole  of  ions  will  transport  96,540  coulombs  of  electricity; 

hence  the  ratio  -      -  will  give  the  fraction  of  the  distance, 
96540 

i  cm.,  which  the  ions  together  have  traversed  in  i  second, 
i.e.,  the  sum  of  their  velocities  in  centimeter-seconds. 
By  Ohm's  law 


where  E  is  the  current  strength,  v  is  the  electromotive 
force,  and  r  is  the  resistance.  Since  v  =  i  in  this  case, 
we  have 


But  —  is  the  conductivity;  hence  instead  of  e  we  may 
use   the   equivalent   conductivity   of   the   solution,    i.e., 


96540    96540* 

This  equation  can  also  be  obtained  directly  from  the 

,  or 


3^4  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

The  molecular  conductivity  of  a  o.oooi  normal 
solution  of  KC1  at  18°  (in  reciprocal  ohms)  is  128.9; 
the  sum  of  the  distances  traversed  by  the  two  ions 
is 

128.9 

=0.001345  cm.  per  second. 


This  total  velocity  is  made  up  of  the  two  single  velocities. 
By  Hittorf's  experiments  the  relative  velocities  of  K*  and 
Cl'  ions  in  KC1  are  in  the  ratio  of  49  to  51.  Hence 
K*  has  a  velocity  of  0.00066  cm.  per  second  and  Cl' 
has  a  velocity  of  0.00069  cm-  Per  second  in  a  o.oooi 
molar  solution  at  18°,  for  a  potential  difference  of  i 
volt. 

Knowing  the  absolute  mobility  of  the  ions  we 
can  thus  calculate  the  equivalent  conductivity  at  infi- 
nite or  any  other  dilution.  We  have,  in  the  two 
cases, 


and 


96540 


where  Ua  and  Uc  are  for  infinite  dilution,  and  expressed 
in   centimeters    per   second   at   a   definite   temperature. 


ELECTROCHEMISTRY.  385 

In  the  table  below  are  some  of  the  absolute  mobilities  as 
given  by  Kohlrausch. 

ABSOLUTE  VELOCITY  OF  THE  IONS  AT  18°  IN  CMS.  PER  SECOND. 


K'         =0.00066 
NH4-     =0.00066 

Na"        =0.00045 
Li*         =0.00036 
Ag'        =0.00057 
Cr2O7"  =  0.000473 

H"      =0.00320 
Cl'     =0.00069 
NO3'  =0.00064 
C1O3'  =  0.0005  7 
OH'  =o.  00181 
Cu"    =0.00031 

It  has  been  possible  to  prove  the  correctness  of  some 
of  these  results  by  experiment,  i.e.,  by  actual  speed 
measurements  on  a  color  boundary  in  a  tube.  Whet- 
ham's  method  for  this  purpose  (Phil.  Trans.,  i893A,  337; 
i895A,  507)  is  as  follows:  If  we  consider  the  boundary- 
line  of  two  equally  dense  solutions  which  contain  a 
common  colorless  ion,  but  which  are  colored  differently, 
and  call  the  salts  AC  and  BC,  then  when  a  current 
passes  through  the  boundary-line  the  C  ions  go  in  one 
direction  and  the  A  and  B  ions  in  the  other.  If  the 
A  and  B  ions  are  the  cathions,  then  the  color  boundary 
will  move  in  the  direction  of  the  current,  and  its  velocity 
will  be  equal  to  the  velocity  of  the  ion  which  causes  the 
change  of  color.  In  this  way  the  velocities  for  Cu", 
Cr2O7",  and  Cl'  were  determined  in  centimeter-seconds. 
In  the  table  on  p.  386  the  values  found  in  this  way  are 
compared  with  those  calculated  by  Kohlrausch. 

For  other  methods  for  determining  the  mobilities  of 
the  ions  see  Masson  (Phil.  Trans.,  IQ2A,  331,  1899); 


3^6  ELEMENTS  OP  PHYSICAL   CHEMISTRY. 

Steele  (Phil.  Trans.,  198,  105,  1902);  Abegg  and  Gans 
(Zeit.  f.  phys.  Chem.,  40,  737,  1902),  and  Denison  (Zeit. 
f.  phys.  Chem.,  44,  575,  1903). 

COMPARISON  OF  ABSOLUTE  IONIC  VELOCITIES. 

T  Whetham,  Kohlrausch, 

by  Experiment.  Calculated. 

Cu" 0.00026  0.00031 

0.000309 
Cl' 0.00057  0.00069 

o .00059 

0.00047 
Cr2O7" 0.00048  0.000473 

0.00046 

91.  The  basicity  of  an  acid. — For  all  binary  organic 
acids  the  formula 


holds  strictly.  For  dibasic  acids  this  is  also  true  when 
the  dissociation  is  not  more  than  50%,  i.e.,  all  dibasic 
organic  acids  under  these  conditions  ionize  as  monobasic 
acids.  In  the  case  of  the  neutral  salts  of  these  acids, 
however,  the  dissociation  into  more  than  two  ions  takes 
place  at  a  much  smaller  dilution.  The  difference  of 
conductivity,  then,  between  two  dilutions  must  be  greater 
for  the  neutral  salt  of  a  polybasic  acid  than  for  one  of  a 
monobasic  one.  Ostwald  has  made  use  of  this  fact  for 
the  determination  of  the  basicity.  He  found  empiri- 
cally that  the  Na  salt  of  a  monobasic  acid  gives  a  differ- 


'ELECTROCHEMISTRY.  387 

ence  in  equivalent  conductivity  at  7  =  32  liters  and 
7  =  1024  liters  of  approximately  10  units,  a  dibasic  one 
of  20  units,  etc.  Hence  to  find  the  basicity  of  an  acid 
we  have  only  to  find  the  equivalent  conductivity  of  its 
Na  salt  at  32  and  1024  liters  dilution;  then  w,  the  basicity 

J 

of  the  acid,  is  given  by  n  =  — . 


10 


VALUES  OF  J  AND  n  FOR  SODIUM  SALTS  OF  ORGANIC  ACIDS. 

Acid.  J  n  =  — 

IO 

Formic  ............................  10.3  i 

Acetic  .............................  9.5  i 

Propionic  ..........................  10.2  i 

Benzoic  ...........................  8.3  i 

Quininic  ..........................  19.8  2 

Pyridin-tricarboxylic  (i,  2,  3)  ........  31  .o  3 

(i,  2,  4)  ........  29.4  3 

Pyridin-tetracarboxylic  ..............  41  .8  4 

Pyridin-pentacarboxylic  .............  50  .  i  5 

92.  The  conductivity  of  neutral  salts.  —  Another  em- 
pirical law  enables  us  to  find  the  equivalent  conductivity 
of  a  neutral  salt  at  one  dilution,  provided  we  know  it 
for  another,  and  the  salt  is  largely  ionized  i.e.,  when 
Av  is  not  very  different  from  A*.  The  relation  observed 
is  as  follows: 


or 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

where  n\  and  n^  are  the  valences  of  the  anion  and  cathion 
respectively,  and  cv  is  a  constant  for  all  electrolytes. 
When  cv  is  known  for  all  dilutions,  and  also  the  terms 
Av,  ni,  and  n%,  we  can  find  the  value  of  A^,  i.e.,  the 
equivalent  conductivity  at  infinite  dilution.  If  we  desig- 
nate (ni-n2-cv)  by  dm  then 


Table  XVI  gives  the  value  of  dv  for  different  dilutions 
and  values  of  n\  -nz&t  25°- 


TABLE  X 


Valence,  n\  .  «a 
I  

.  .    ii 

8 

^256 

6 

^512 

4 

^1024 
•7 

2    

.  .     21 

16 

12 

8 

6 

3O 

23 

17 

I  2 

8 

4  *  .  . 

.  .     42 

23 

16 

10 

c 

20 

21 

13 

6.  . 

.     60 

48 

36 

2? 

16 

This  behavior  may  be  summed  up  in  words  as  follows : 
The  decrease  0}  equivalent  conductivity  is  roughly  Con- 
stant jor  salts  oj  the  same  type;  and  the  decrease  in  equiva- 
lent conductivity  jor  salts  oj  different  types  is  proportional 
to  the  valences  oj  the  ions. 

The  relation  of  ionization  to  dilution  in  those  cases 
where  the  Ostwald  dilution  law  cannot  be  applied  has 
already  been  considered  (see  pp.  293-301  and  305-316). 

93.  The  ionization  of  water. — The  ions  of  water  are 
H*  and  OH',  i.e.,  to  a  very  slight  degree  it  is  a  binary 


ELECTROCHEMISTRY  389 

electrolyte.    The  specific   conductivity  K  of   a  specially 
purified  water  was  found  by  Kohlrausch  to  be 

0.014 -io~6  at    o°  C. 
o.o4o-io~6at  1 8° 
0.055 -io~6  at  25° 
0.084  •  io~6  at  34° 
0.17   -io~6  at  50° 

One  mm.  of  this  water  at  o°  has  a  resistance  equal 
to  that  of  a  copper  wire  of  the  same  section  and 
40,000,000  kilometers  long,  i.e.,  which  is  long  enough 
to  go  around  the  earth  a  thousand  times. 

From  this  conductivity  the  degree  of  dissociation  is 
easily  determined.  One  mole  of  H'  ions  has  a  con- 
ductivity equal  to  318  units  (p.  381),  while  i  mole  of 
OH'  ions  has  one  equal  to  174;  hence  the  maximum 
molecular  conductivity  of  water  should  be  AM  =318  +  1 74 
=  492,  if  completely  dissociated.  That  of  a  liter  be- 
tween electrodes  i  cm.  apart  at  18°  is  lo3  times  as 
large  as  0.04 Xio~6,  i.e., 

0.04  X  io~6  X  lo3 =0.04  X  io~3, 
hence 

0.04  X  i  o~3 


492 


=o.8Xio~7; 


39°  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

which  is  the  concentration  of  H'  and  OH'  ions  in  moles 
per  liter  at  18°,  or  there  are  17  grams  of  OH'  and  i 
gram  of  H'  ions  in  12,000,000  liters  of  water. 

It  is  to  be  remembered  here  that  the  492  would  be  the 
value  if  i  mole  of  H'  ions  and  i  mole  of  OH'  ions  were 
present  together  between  electrodes  i  cm.  apart  and  of 
any  cross-section,  so  long  as  they  will  contain  between 
them  the  amount  of  solution.  Since  we  use  water  as  a 
solvent,  the  equivalent  conductivity  is  usually  calculated 
for  a  liter,  and  not,  as  it  might  be  assumed  by  definition, 
1  8  grams.  The  calculation  was  similar  in  determining 
the  ionic  product  of  H*  and  OH'  ions  (p.  313),  where  Sn2o 

1000^ 
was  equal  to  —  z-K,  i.e.,  55«5A. 

Io 

94.  The  temperature  coefficient  of  conductivity.  —  Ac- 

cording to  Kohlrausch,  the  conductivity  varies  with  the 
temperature  as  follows: 


where  ft  'is  the  temperature  coefficient,  or  the  change 
in  conductivity  for  i°  C.  We  can  find  /?  by  experiment 
from  the  equation 


Substances  with  small  molecular  conductivities  usually 
have  large  temperature  coefficients.     Most  of  the  strongly 


ELEC  TROCHE  MIS  TRY.  391 

dissociated  salts  have  a  value  for  /?  equal  to  0.025,  i.e.,  the 
equivalent  conductivity  changes  2\%  for  each  degree. 

This  temperature  coefficient  depends  upon  several 
factors,  for  the  internal  friction  of  the  solvent  changes, 
as  well  as  the  degree  of  dissociation  and  the  velocity 
of  migration.  It  is  only  possible  to  obtain  a  negative 
coefficient  when  the  dissociation  decreases  enough  to 
more  than  compensate  for  the  increase  in  the  velocity 
due  to  the  diminished  friction;  consequently  all  such 
substances  must  have  negative  heats  of  dissociation. 

The  equivalent  conductivity  at  infinite  dilution  also 
increases  with  the  temperature,  so  it  must  not  be  im- 
agined that  the  ionization  increases  in  this  way.  The 
examples  already  given  (p.  299)  will  serve  to  show  how 
a  varies  with  the  temperature. 

95.  The  conductivity  of  difficultly  soluble  salts.  —  If 
the  saturated  solution  of  any  insoluble  salt  is  so  dilute 
that  we  may  assume  complete  dissociation,  Aw  =AV,  we 
have 


^solution  ~~ 

and 

Av=Ago  =*XioooF, 
i.e., 


392  .  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

and 

i      *Xio3 
c  =  yr  =  —  --  moles  per  liter. 

This  method  has  been  used  by  Hollemann  *  and 
Kohlrausch  and  Rose  f  with  very  good  results.  In 
this  way  a  number  of  solubilities  have  been  determined. 
Thus 

i  part  BaSO4  dissolves  in  429,700  parts  H^O  at  i6®.i  C., 
i  part  SrSOd  dissolves  in  10,070  parts  H2O  at  24°.  2  C., 
i  part  BaCOs  dissolves  in  45,566  parts  H^O  at  24°.3  C., 
i  part  SrCOa  dissolves  in  91,468  parts  H^O  at  24°.3  C. 

BaSO4  is  of  course  so  insoluble  that  it  is  impossible 
to  find  its  term  A^  by  experiment.  Its  value,  however, 
can  be  found  from  the  general  equation 


where  M  and  Mi  represent  metals  and  X  and  Xi  the 
negative  radicals. 
Thus  from 

4,,  for  JBaCl2,  =115] 
4,,  forKCl,         =122     18° 


*  Zeit.  f.  phys.  Chem.,  12,  125,  1893.      f  Ibid.,  234,  1893. 


ELECTROCHEMISTRY.  393 

we  have 

A^,  for  BaSO4,  =  II5  +  128  —  122  =  121. 

This  value  could  also  be  found  from  the  table  illus- 
trating Kohlrausch's  law  of  the  independent  migration 
of  the  ions,  of  which  it  is  an  application,  or  from  that 
on  p.  381.  It  must  be  borne  in  mind  here  that  this 
method  is  only  applicable  when  we  can  safely  assume  Av 
for  a  saturated  solution  to  be  the  same  as  A^.  And 
further,  it  is  limited  to  those  substances  which  are  soluble 
enough  to  give  a  value  of  K  which  differs  appreciably 
from  .that  for  the  water  used.  Silver  iodide  solutions  in 
this  respect  seems  to  be  about  on  or  just  beyond  this 
limit. 

96.  The  dielectric  constant  and  dissociating  power. 
Other  solvents  than  water. — According  to  J.  J.  Thorn  sen 
and  Nernst,  there  is  a  relation  between  the  dielectric 
constant  of  the  solvent  and  its  power  of  dissociating 
dissolved  substances.  The  dielectric  constant  is  deter- 
mined by  placing  the  substance  between  two  plates 
between  which  there  is  a  certain  potential  difference, 
and  measuring  the  loss  of  potential.  Air  is  used  first 
between  the  plates,  and  then  the  substance  to  be  de- 
termined. The  factor  which  we  measure  is  the  specific 
inductive  capacity  of  the  liquid  for  electricity,  referred 
to  air  as  a  standard.  The  dielectric  constants  of  a 
number  of  solvents  are  given  in  the  following  table. 


394  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

Benzene 2 . 24 

Ether 4.25 

Amyl  alcohol 16.05 

Ethyl  alcohol,  99.8% 25 .8 

Water 79 . 6 

As  will  be  observed,  the  constant  for  water  is  higher 
than  for  any  of  the  other  solvents,  and  its  dissociating 
power  is  also  the  greatest. 

The  latest  theory  as  to  the  relation  between  the  dielec- 
tric constant  and  dissociating  power  is  by  Briihl.*  The 
dielectric,  i.e.,  the  solvent,  acts  just  as  the  glass  in  a 
Leyden  jar  and  prevents  the  ions  from  neutralizing  their 
charges.  The  greater  the  dielectric  constant,  then,  the 
greater  the  amount  of  free  ions  which  may  exist  in  the 
solution.  This  connection  between  the  dielectric  constant 
and  dissociation,  as  found  by  electrical  and  other  means, 
may  be  considered  as  very  strongly  confirmatory  of  the 
theory  of  dissociation  into  charged  components  or  ions. 

The  electrical  behavior  of  substances  in  many  non- 
aqueous  solvents  is  very  different  from  that  in  water, 
and  it  is  difficult  from  our  present  knowledge  to  see  how 
they  can  be  governed  by  the  same  laws.  For  examples  in 
some  cases  the  equivalent  conductivity  does  not  attain 
a  maximum  constant  value,  but  after  attaining  a  maxi- 
mum decreases  with  further  dilution;  and  the  degree  of 
ionization  determined  electrically  does  not  in  all  cases 
agree  with  that  determined  by  the  boiling-point,  for 

*  Zeit.  f.  phys.  Chem.,  30,  i,  1899. 


ELECTROCHEMISTRY.  395 

instance.  Further  than  this,  the  conductivity  cannot  in 
all  cases  be  said  to  be  the  sum  of  two  independent 
values,  i.e.,  of  those  of  the  ions. 

We  cannot  but  infer,  then,  from  our  present  knowl- 
edge that  the  process  of  solution  is  quite  different  for 
these  than  for  water.  Neither  the  Ostwald  nor  any 
other  form  of  a  dilution  law  seems  to  hold  in  these  cases. 
In  other  words,  the  electrical  behavior  of  these  is  in- 
explicable as  yet,  and  until  the  processes  taking  place 
in  such  systems  is  better  understood  little  of  our  data 
can  be  generalized.  There  is  one  exception  to  all  this, 
however.  Franklin  and  Kraus  (Am.  Chem.  J.,  23,  277, 
1900)  have  found  that  dilute  binary  electrolytes  in 
liquefied  ammonia-gas  give  an  approximately  constant 
value  for  the  Ostwald  dilution  law,  at  least  a  value 
which  very  much  more  nearly  constant  than  that  ob- 
tained in  a  water  solution.  For  concentrated  solutions 
in  the  same  solvent,  see  Franklin  and  Kraus  (J.  Am. 
Chem.  Soc.,  27,  191,  1905). 

The  behavior  of  metallic  sodium  in  liquefied  am- 
monia (Cady,  J.  Phys.  Chem.,  i,  707,  1897,  and  Franklin 
and  Kraus,  1.  c.,  p.  306)  is  very  peculiar,  for  the  con- 
duction is  apparently  more  metallic  than  electrolytic. 
No  products  of  decomposition  are  observed  and  there 
is  no  polarization,  while  the  conductivity  is  exceedingly 
high. 

Among  the  tables  at  the  end  of  this  book  will  be  found 


396  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

one  giving  the  conductivities  of  substances  in  various 
non-aqueous  solvents,  from  which  an  idea,  at  any  rate, 
may  be  obtained  as  to  their  relations  toward  the  electric 
current. 

97.  The  conductivity  of  mixtures  of  substances  having 
an  ion  in  common.— The  conductivity  of  a  solution  con- 
taining two  substances  with  an  ion  in  common  can  be 
readily  shown  to  be  given  by  the  formula 


K = 


where  ^i  and  V2  are  the  volumes  of  the  two  simple  solu- 
tions which  are  mixed,  i.e.,  Vi+v2  is  the  total  volume 
of  solution  containing  n\+n2  equivalents  of  the  two 
substances.  a\  and  a2  are  the  degrees  of  ionization  of 
the  substances  in  the  mixture,  A^\  and  A^2  are  the 
equivalent  conductivities  of  the  two  substances  in  a 
simple  solution,  and  p  is  the  ratio  of  the  volume  of  the 
mixture  to  the  sum  of  the  constituent  volumes. 

For  the  freezing-point  of  such  a  mixture,  in  an  analo- 
gous way,  we  have 

A  =[MiNi(i  +ai(mi  -  1))  +  ^2^2(1  +a2(m2  -  1))], 


where  m  is  the  number  of  moles  of  ions  produced  from 
one  original  mole,  M  i  and  M2  are  the  molecular  depres- 


ELEC  TROCHEMIS  TRY.  397 

sions  (p.  183)  in  i  liter  of  the  simple  solutions  and  N\ 
and  N2  are  the  number  of  moles  per  liter  of  each. 

For  substances  for  which  the  Ostwald  dilution  law 
holds  there  is,  naturally,  no  difficulty  in  determining 
the  terms  a\  and  #2  (p-  283) ;  and  for  strong  electrolytes 
it  is  also  possible,  although  the  process  is  not  always 
so  simple. 

For  illustrations  of  the  use  of  these  formulas  and 
the  calculation  of  the  values  a\  and  a 2  the  reader  must 
be  referred  elsewhere,*  for  it  would  lead  us  too  far  to 
consider  them  in  detail  here.  It  will  be  observed,  how- 
ever, that  it  is  assumed  in  the  above  formula  that  A^\ 
and  AM2  are  not  altered  by  the  mixing,  a  supposition 
which  probably  holds  only  for  solutions  weaker  than 
0.5  normal. 

Noyes  and  Kohr  (J.  Am.  Chem.  Soc.,  24,  1145-1146, 
1902)  give  an  example  of  the  calculation  of  the  degree  of 
ionization  of  such  substances  when  mixed,  which  is 
based  upon  another  principle  and  somewhat  simpler  than 
the  other  methods.  The  principle  upon  which  this  de- 
termination is  based  is  as  follows:  The  dissociation  of 
a  substance  in  the  presence  of  another  with  an  ion  in 
common  is  that  which  the  conductivity  measurements 
show  it  to  have  when  it  is  alone  present  at  a  concentra- 


*  See  MacGregor  (Trans.  Nova  Scotia  Inst.  of  Science,  9,  101,  1896); 
Barnes  (ibid.,  10,  124,  1900),  and  Archibald  (ibid.,  9,  291,  1898; 
also  Trans.  R.  S.  of  Can.  (2),  3,  69,  1897). 


398  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

tion  of  its  ions  equal  to  the  square-root  of  the  product  of 
the  concentration  of  its  ions  in  the  solution  of  the  mixed 
salts.  This  principle  was  demonstrated  by  solubility 
experiments  (Noyes  and  Abbott,  Zeit.  f.  phys.  Chem., 
16,  138,  1895). 

It  was  found  there  that  for  those  substances  which  do 
not  follow  the  Ostwald  dilution  law  the  concentration 
of  the  un-ionized  part  of  a  salt  has  the  same  value, 
when  the  product  of  the  concentrations  of  its  ions  has 
the  same  value,  whatever  may  be  the  two  separate  factors 
of  that  product.  In  a  mixture  of  potassium  chloride  and 
potassium  hydroxide,  for  example,  where  the  former 
is  0.0033  m°lar  and  the  latter  0.35,  we  first  assume 
the  ionization  to  be  85%  for  each  (as  an  approximation). 
The  product  CK-  Xccl/.  then  has  the  approximate  value, 
(0.35  +  0.0033)  X  0.85X0.0033X0.85.  The  square  root 
of  this  quantity  (or  0.030)  is  then  85%  of  the  concen- 
tration (0.0355)  of  pure  potassium  chloride  at  which  the 
degree  of  dissociation  is  the  same  as  it  is  in  the  solution 
of  the  mixed  potassium  salts.  The  degree  of  dissociation 
is  then  to  be  ascertained  from  a  plot  of  the  molar  con- 
ductivity values.  For  potassium  chloride  in  this  way  we 
find  a  to  be  0.90;  and  by  the  same  procedure  a  for  the 
KOH  is  found  to  be  84%. 

In  mixtures  of  substances  without  an  ion  in  common, 
the  same  general  formula  for  the  conductivity  and  freez- 
ing-point (p.  396)  is  also  to  be  employed.  Here,  how- 


ELECTROCHEMISTRY.  399 

ever,  there  may  be  a  reaction  between  the  substances 
which  would  give  rise  to  a  new  substance.  In  such  a 
case,  this,  also,  must  be  considered,  and  treated  just  as  the 
others.  All  these  things  also  hold  for  mixtures  of  more 
than  two  substances,  although  in  such  cases  the  calcula- 
tion is  considerably  more  complicated  (see  MacGregor, 
Trans.  Roy.  Soc.  Can.  (2),  2,  65,  1896). 

C.    ELECTROMOTIVE  FORCE. 

98.  Determination  of  the  electromotive  force. — In  all 

the  cases  we  are  to  consider  in  this  section  it  is  neces- 
sary that  the  electromotive  force  be  measured  by  aid 
of  the  compensation  method,  i.e.,  that  as  little  actual 
current  be  taken  from  the  cell  as  is  possible. 

As  the  basis  of  such  methods  is  a  known  and  con- 
stant electromotive  force,  the  standard  cell  is  an  ex- 
ceedingly important  part  of  the  measurement.  The 
standard  cells  in  general  use  in  this  work  are  of  two 
kinds,  one  of  which  has  a  fixed  constant  value,  while 
the  other  may  be  made  up  in  such  a  way  as  to  give 
one  of  several  values. 

The  former  one  is  the  Clark  cell,  Fig.  17,  which  con- 
sists of  amalgamated  Zn  in  a  paste  of  ZnSC>4  and  Hg 
in  a  paste  of  Hg2SO4.  This  cell  gives  an  E.M.F.  equal 
to 

1.4328  — o.oong(/  — 15)  —  0.000007  (/  —  i  5)2. 


4oo 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


Another  standard  cell  in  common  use  is  the  cadmium 
element    (Weston).    This   is   made   up   with   electrodes 


FIG.  17. 

of  mercury  and   cadmium   amalgam   according  to   the 
following  scheme,  the  Hg  being  the  positive  pole: 

Hg-Hg2SO4  +  CdSO4 

paste 

sat.  CdSC>4  +  crystals 
CdSC>4  —  Cd  amalgam. 


paste 


The  electromotive  force  of  this  cell  is 

1.0186  +  0.00004(20— /)  volts. 

The  Helmholtz  element,  Fig.  18,  consists  of  amalga- 
mated  Zn   in   ZnCb  solution    (sp.   gr.  =1.409   at    15°), 


ELECT ROCHEM1S  TRY.  401 

calomel,  and  Hg.  This  gives  an  E.M.F.  approximately 
equal  to  i  volt,  but  its  exact  value  depends  upon  the 
sp.  gr.  of  the  ZnCl2  solution  so  that  it  must  always  be 
measured  by  comparison  with  a  Weston  or  Clark  ele- 


Calomel 


FIG.  18. 

ment.  Its  great  advantage  is  its  very  small  tempera- 
ture coefficient,  which  at  low  temperatures  is  negligible.* 
99.  Types  of  cells. — Cells  are  either  reversible  or 
non-reversible,  according  as  the  process  may  be  reversed 
or  not.  A  reversible  cell  can  always  be  put  in  its  orig- 
inal condition  again  by  sending  a  current  through  it  in 
the  opposite  direction,  while  with  a  non- reversible  cell 
this  is  not  possible.  A  typical  reversible  cell  is  the 
Daniell,  where  we  have  Cu  in  CuSC>4  and  Zn  in  ZnSO4. 


*  For  further  details  of  these  measurements,  cells,  etc.,  see  Ostwald's 
Physico-chemical  Measurements,  Macmillan. 


402  ELEMENTS  OP  PHYSICAL  CHEMISTRY. 

The  Zn  dissolves  and  Cu  is  precipitated  during  the 
action,  but  by  passing  a  current  through  it  from  Cu 
to  Zn  the  Zn  is  precipitated,  while  Cu  is  dissolved,  until 
the  original  state  is  again  reached. 

A  non-reversible  cell,  on  the  other  hand,  is  any  one 
from  which  a  gas  is  given  off,  for  the  passage  of  the 
current  cannot  cause  the  gas  to  recombine.  Here  we 
shall  consider  principally  the  reversible  type. 

100.  The  chemical  or  thermodynamical  theory  of  the 
cell. — The  simplest  theory  of  the  action  of  a  reversible 
cell  was  advanced  by  Helmholtz  and  Thomson.  Accord- 
ing to  this,  the  chemical  energy  of  the  process  is  trans- 
formed completely  into  electrical  energy.  This  theory  in 
the  form  in  which  it  was  advanced,  however,  has  proven 
to  be  incorrect,  for  it  holds  only  for  the  Daniell  cell.  The 
explanation  of  the  variation  in  the  case  of  other  cells 
was  offered  by  Gibbs  and  Helmholtz  independently,  who 
proved  it  to  be  due  to  the  temperature. 

Assume  a  reversible  primary  cell  in  which  the  amount 
of  heat  q  is  liberate'd  or  absorbed  when  one  equivalent 
of  the  ions  has  been  carried  from  one  side  to  the  other. 
Imagine  this  cell  in  a  temperature-bath  so  that  the  cell 
remains  at  constant  temperature, — i.e.,  if  heat  is  ab- 
sorbed it  is  replaced,  or  if  heat  is  liberated  it  is  removed 
— thus  preventing  q  from  causing  a  change  in  temperature. 

If  TT,  the  E.M.F.  of  this  cell,  is  just  compensated  by 
— TT,  the  process  will  be  in  equilibrium.  For  two  ener- 


ELECTROCHEMISTRY.  40  3 

gies  in  equilbirum,  however,  we  have  found  the  general 
equation  (p.  6) 


where  c  and  i  are  the  factors  of  one  kind  of  energy  and 
c\  and  i\  those  of  the  other  kind.  In  our  case  the  two 
kinds  of  energy  are  electrical  and  thermal,  and  we  have 


or 


which  is  the  change  in  the  E.M.F.  of  a  cell  caused  by 
the  absorption  or  liberation  of  q  cals.  during  the  reac- 
tion. 

If  now  we  pass  96,540  coulombs  of  electricity  through 
the  cell,  during  which  to  keep  the  temperature  constant 
it  is  necessary  to  supply  q  cals.,  the  electrical  energy 
supplied  must  be  equal  to  the  chemical  energy  plus  the 
electrical  energy  due  to  the  heat  q.  We  have,  then,  when 
all  terms  are  expressed  in  the  same  kind  of  units, 


404  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

or,  since  Ee= 


i.e.,  the  actual  E.M.F.,  TT,  is  only  equal  to  that  calcu- 
lated from  the  chemical  energy  when  dn=o,  i.e.,  when 
the  E.M.F.  does  not  change  with  the  temperature.  Other- 

wise TT  is  smaller  or  greater  than  —  -  according  as  T-^ 

£Q  dl 

is  negative  or  positive. 

The  Daniell  cell  has  such  a  small  temperature  coeffi- 
cient that  the  electrical  energy  is  nearly  equal  to  the 
chemical  energy.  For  other  cells  it  is  possible  to  deter- 
mine the  temperature  coefficient,  and  thus  from  the 
chemical  energy  to  calculate  the  E.M.F.  of  the  cell. 
The  results  by  experiment  and  calculation  agree  in  most 
cases  within  the  experimental  error,  showing  that  this 
conception  of  the  process  is  correct.  We  see,  then,  that 
the  amount  of  heat  generated  by  the  chemical  process 
is  no  criterion  of  the  amount  of  electrical  energy  involved, 
for  heat  which  is  absorbed  from  the  environment  may 
also  be  transformed  into  electrical  energy,  and  in  other 
cases  less  heat  may  be  transformed  into  electricity  than 
is  given  up  by  the  chemical  process. 

An  example  of  the  application  of  this  formula  is  given 
by  the  Grove  gas  cell.  Here  71  =  1.062  and  £  =  34200, 

34200        dn  dn 

hence      1.062=  -      -  +  T~^,     i.e.,     T-^=  -0.4187,  m 
23110        dT  dT 


ELECTROCHEMISTRY.  4°5 

place  of  —0.416,  as  found  by  experiment.  The  value 
23110  here  is  obtained  from  96540X0.2394X^  =  23110/1 
cals.,  and  is  the  work  in  calories  necessary  for  the 
deposition  of  i  gram  equivalent  of  any  substance  a-t 
it  volts. 

101.  The  osmotic  theory  of  a  cell. — Considering  a 
cell  from  the  standpoint  of  our  conclusions  respecting 
the  nature  of  electrolytes  in  solution,  it  is  possible  to 
see  more  clearly  into  the  cause  of  the  rise  of  a  difference 
in  potential  between  two  solutions  or  a  metal  and  a 
solution. 

Assume  that  we  have  two  solutions  in  contact,  and 
that  they  contain  the  same  monovalent  ions  in  different 
concentrations.  The  difference  of  potential  existing  on 
their  boundary  can  be  calculated  by  aid  of  the  following 
process  of  reasoning:  If  Ua  and  Uc  are  the  mobilities  of 
the  respective  ions,  then,  by  the  passage  of  96,540  cou- 
lombs of  electricity,  the  following  changes  take  place: 
If  the  current  enters  on  the  concentrated  side  and  passes 

through  both  solutions,  -— — ^—~  gram  equivalents  (moles) 

U c  +  U  a 

of  positive  ions  will  go  from  the  concentrated  to  the  dilute 

solution,  and  during  the  same  time  Jr     "     gram  equiva- 

UC~T  U  a 

lents  (moles)  of  negative  ions  will  go  from  the  dilute  to  the 
concentrated  side.  Let  p  be  the  osmotic  pressure  of  the 
positive  and  negative  ions  in  the  concentrated  solution, 


406  ELEMENTS  OP  PHYSICAL   CHEMISTRY. 

and  p'  that  of  those  in  the  dilute  solution.  The  maxi- 
mum work  to  be  done  by  the  process,  then,  will  be 
(P-  225)> 


and  this  must  be  equal  to  £QX  for  the  process  going  in 
this  way,  i.e.,  to  the  electrical  work  done  at  the  surface 
of  contact  of  the  two  solutions.  Since  R  is  in  calories 
this  value  will  also  be  expressed  in  calories.  To  trans- 
form it  into  electrical  units  it  is  only  necessary  to  divide 
through  by  23,110.  This  relation  was  derived  and  tested 
experimentally  by  Nernst  (Zeit.  f.  phys.  Chem.,  4,  1295 
1888),  who  found  it  to  hold  true  within  the  experimental 
error. 

If  Ue  is  greater  than  Ua  we  could  expect,  then,  that 
a  weaker  solution  of  an  acid  or  base,  when  superim- 
posed on  a  stronger  one,  would  show  the  polarity  of  the 
faster  moving  ion.  In  other  words  in  diffusing  through 
the  solution  the  faster  ion  would  impart  its  polarity  to 
the  weaker  solution.  With  acids,  indeed,  we  do  observe 
that  the  dilute  solution  is  positive,  while  with  bases  it  is 
negative  ;  and  this  is  in  accord  with  our  previous  observa- 
tions that  H*  and  OH7  are  the  ions  possessing  the  greatest 
mobility  or  velocity.  This  process  does  not  result  in 
a  complete  separation  of  the  ions  nor  even  in  more  than 


ELECTROCHEMISTRY.  4°  7 

a  very  slight  one,  for  it  must  be  remembered  that  the 
two  kinds  of  electricity  attract  one  another,  the  conse- 
quence of  which  is  that  the  speed  of  the  faster  moving 
ions  is  decreased,  while  that  of  the  slower  is  increased. 

This  same  method  of  reasoning  will  also  enable  us 
to  understand  the  action  of  the  arrangement 

Concentrated       amal-  I  Water    solution   of    a  I  Dilute  amalgam  of  the 
gam  of  the  metal  M.  I     salt  of  the  metal  M  .  \     metal  M. 

In  this  case  the  passage  of  96,540  coulombs  will  cause 
i  gram  equivalent  of  the  metal  M  to  go  from  the  con- 
centrated side  to  the  dilute,  and,  if  there  are  n  equiva- 
lents to  the  mole,  the  maximum  work  per  equivalent 
will  be 


where  c\  and  €2  are  the  concentrations  of  the  metal  M 
in  the  amalgams.  Again,  here,  if  this  expression  is  so 
transformed  as  to  give  electrical  units  instead  of  calories, 
by  division  by  23,110  (i.e.,  96,540X0.2394),  it  will 
give  the  value  of  TT  the  electromotive  force  of  the  cell 
operating  in  this  way. 

And  the  same  is  true  for  cells  with  electrodes  of  the 
same  soluble  metal  and  two  concentrations  of  a  solution 
of  a  salt  or  salts  of  that  metal.  Consider,  for  example, 
the  cell 


408  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


Cu 


dilute 
CuS04 


concentrated 
CuSO4 


Cu. 


By  passing  a  current  through  this  cell  in  the  direction 
of  the  arrow  the  following  changes  will  take  place: 

1.  For   each   96,540   coulombs   of  .electricity    i    gram 
equivalent  of  copper  will  dissolve  from  the  electrode  in 
the  dilute  solution,   i.e.,  will  be  transformed  from  the 
metallic  to  the  ionic  state. 

2.  At  the  boundary  of  the  two  solutions  the  process 
described  above  will  take  place;  and 

3.  One  gram  equivalent  of  Cu"  ions  will  be  deposited 
from  the  concentrated  solution  upon  the  electrode. 

As  the  result  of  processes  (i)  and  (3)  i  gram  equiva- 
lent of  Cu"  ions  will  go  from  the  concentrated  to  the 
dilute  solution.  The  maximum  work  of  this  process, 
then,  for  each  96,540  coulombs  will  be 

RT 


where  n  is  the  valence  of  the  metal  and  p\  and  p2  are  the 
osmotic  pressures  of  the  Cu"  ions  in  the  two  solutions. 

By  the  second  process  (contrast  with  direction  of  cur- 
rent in  the  case  above)  we  have  as  the  work 


n 


ELECTROCHEMIS  TR  Y  409 

This  second  value  is  usually  so  small,  when  compared 
to  the  other,  that  it  may  generally  be  neglected,  and  the 
two  differences  of  potential  between  the  electrodes  and 
the  solutions  measured  separately. 

Neglecting  the  difference  of  potential  between  the 
liquids,  then,  we  have 

RT         p, 
n=  —  log,  — , 

or,  including  the  potential  difference  between  the  solutions, 

RT          2Un 


i.e.,  the  sum  of  the  two,  where  R  must  be  given  in  electri- 
cal units,  if  T.  is  to  be  expressed  in  volts. 

TT>  rri  , 

By  separating  the  amount  of  work  —  log,  —  into  two 

n          p2 

portions,  so  that  each  will  give  the  maximum  work  at 
an  electrode,  we  can  write 

RT.      P     RT 


where  P  is  a  constant  for  any  one  metal  at  the  same 
temperature,  in  the  same  solvent,  and  is  called  the  electro- 
lytic solution  pressure.  The  portions  containing  the  ex- 

p 

pressions  log,  —  will  then  give  the  difference  of  potential 


410  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  copper  in  dilute  copper  sulphate  and  copper  in  con- 
centrated copper  sulphate. 

The  electrolytic  solution  pressure  assumed  to  exist  in 
this  way  is  analogous  to  the  ordinary  solution  pressure 
which  we  have  already  considered,  except  in  this  case  it 
is  for  the  solution  of  a  metal  or  any  other  element,  i.e., 
is  the  pressure  with  which  element  tends  to  attain  the 
ionic  state. 

When  we  consider  a  stick  of  metal  in  a  solution  of 
one  of  its  salts,  we  have,  if  P  is  the  electrolytic  solution 
pressure  and  p  is  the  osmotic  pressure  of  the  metal  ions 
in  the  solution,  three  possibilities: 

1.  P=p. 

2.  P>p. 

3.  P<p. 

The  behavior  in  these  three  cases  we  shall  consider. 

1.  P=p.    We  shall  observe  no  action,  since  the  two 
opposing  pressures  are  equal  and  consequently  cancel. 

2.  P>p.    This    is    the   case  with   Zn.     Ions  of   Zn" 
will  go  from  the  electrode  into  the  solution,  taking  with 
them  positive  electricity.     This  will  take  place  only  to 
a  very  slight  degree,  for  it  would  increase  the  positive 
charge  of  the  solution.    These  positive  ions  around  the 
electrode  will  then  be  attracted  back  to  the  plate,  which 
is  negatively  charged  owing  to  the  loss  of  the  positive  ions. 
We  have  then,  finally,  the  metal  negative  against  the  solu- 


ELECTROCHEMISTRY.  411 

tion,  and  the  electrode  surrounded  by  what  is  known  as 
a  Helmholtz  double  layer,  in  this  case  a  layer  of  positive 
ions  on  a  negative  plate.  If  positive  electricity  is  given 
now  to  the  metal,  the  double  layer  is  broken  up  and 
more  Zn  will  go  into  solution  in  the  ionic  form,  but  as 
soon  as  the  current  is  stopped  the  double  layer  will  again 
be  formed. 

3.  P<p.  This  is  true  for  Cu.  A  stick  of  copper 
in  a  solution  of  one  of  its  salts  will  have  ions  of  Cu"  from 
the  solution  deposited  upon  it,  since  the  pressure  toward 
the  Cu  is  greater  than  that  away  from  it.  These  ions 
take  positive  electricity  with  them,  consequently  the 
metal  is  positive  against  its  solution.  The  double  layer 
will  also  be  found  here,  for  the  positive  charge  on  the 
metal  will  attract  the  negative  ions  from  the  solution. 
In  this  case,  however,  the  double  layer  will  be  broken 
up  when  electricity  is  conducted  away  from  the  Cu,  so 
that  the  Cu"  ions  will  again  precipitate  and  continue 
to  do  so  until  p=P,  when  all  action  will  cease.  Since 
i  gram  equivalent  of  ions  carries  with  it  96,540  coulombs 
of  electricity,  the  amount  of  metal  going  in  or  out  of 
solution  may  be  infinitesimal  and  still  cause  the  produc- 
tion of  a  measurable  amount  of  electricity. 

The  relations  of  Cu  and  Zn  in  their  respective  solu- 
tions are  shown  in  Fig.  19. 

The  double  layer  is  formed  on  both.  Until  the  Cu 
and  Zn  are  connected  by  a  wire  the  equilibrium  as  it 


412 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


is  will  be  undisturbed.  When  they  are  connected  a 
current  will  go  through  the  wire  from  Cu  to  Zn,  i.e., 
Zn"  ions  will  go  into  the  solution,  depositing  an  equiva- 
lent amount  of  Cu"  ions  upon  the  copper  plate.  The 
electromotive  force  will  then  depend  on  the  difference 
between  the  two  values  of  P,  if  p  is  the  same  on  both 


FIG.  19. 


FIG.  20. 


sides,  and  the  amount  of  current  will  depend  upon  the 
number  of  gram  equivalents  of  Zn  dissolved  and  of  Cu 
precipitated. 

The  value  of  the  osmotic  counter-pressure,  it  will  be 
seen,  regulates  the  behavior  of  the  metal,  i.e.,  indicates 
whether  it  will  dissolve  or  not.  Thus  Cu  and  Ag  dis- 
solve when  KCN  is  added  to  their  solutions;  this  means 
that  the  osmotic  pressure  of  the  ions  of  Ag-  or  Cu"  in 
the  solution  is  smaller  than  the  solution  pressure,  for  the 


ELECTROCHEMISTRY.  4*3 

Cu"  or  Ag'  ions  combine  with  those  of  CN'  to  form 
complex  ions  and  are  thus  removed  from  the  system. 

The  presence  of  a  layer  of  ions  around  the  charged 
metal  has  been  proven  by  Palmaer  (Wied.  Ann.,  28,  257, 
1899)  by  an  arrangement  shown  in  Fig.  20. 

Drops  of  mercury  are  allowed  to  fall  into  a  weak 
solution  containing  mercurous  ions,  metallic  mercury 
being  in  the  bottom  of  the  tube  containing  the  solution. 
If  now  the  double  layer  theory  is  true,  the  drops  of  Hg 
as  they  form  should  have  the  electricity  of  the  Hg^  ions 
deposited  upon  them,  and  these  positively  charged  drops 
should  then  attract  the  negative  ions,  forming  on  each  a 
double  layer.  When  such  a  drop  reaches  the  mercury 
at  the  bottom  it  will  unite  with  that,  forming  once  more 
ions  of  Hg2  and  releasing  those  of  the  negative  radical, 
and  the  concentration  of  mercury  salt  will  be  greater  at 
the  bottom  than  at  the  top.  Palmaer's  experiments 
showed  this  difference  in  concentration  to  actually  exist. 

Experiment  has  shown  that  the  metals  Na,  K .  . .  , 
etc.,  up  to  Zn,  Cd,  Co,  Ni,  and  Fe  are  always  negative 
against  their  solutions,  i.e.,  P>  p. 

The  noble  metals,  on  the  contrary,  are  positive  against 
their  solutions,  although  in  some  few  cases  it  is  possible 
to  get  a  solution  in  which  P>p.  In  general,  though, 
for  the  noble  metals  P<p. 

A  negative  element  has  exactly  the  same  action  except 
that,  in  general,  as  far  as  is  known,  P>  p.  Here,  although 


414  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

P>p,  the  electrode  is  positive  against  the  solution,  for 
the  negative  ions  formed  from  the  electrode  leave  posi- 
tive electricity  behind. 

In  'general,  the  electrolytic  solution  pressure  depends 
upon  the  temperature,  the  nature  of  the  solvent,  and 
the  concentration  of  the  substance  in  the  electrode  (see 
p.  422). 

102.  Mathematical  expression  of  the  osmotic  theory  of 
the  cell.  —  It  is  possible  from  the  above  conception  of 
solution  pressure  to  derive  the  formula  giving  the  value 
of  the  E.M.F.  of  a  single  electrode  of  any  metal  in  a  solu- 
tion of  one  of  its  salts.  This  will  depend  upon  the  osmotic 
pressure  of  the  metal  ions  working  in  the  one  direction 
and  upon  the  electrolytic  solution  tension  working  in 
the  other.  Since  the  ions  go  from  one  pressure  to  the 
other,  a  certain  amount  of  work  is  done  and  this  amount 
must  always  be  the  same  no  matter  in  what  form  it 
appears,  i.e.,  whether  electrical  or  osmotic. 

When  i  mole  of  ions  is  transferred  from  the  pressure 
P  to  the  pressure  p,  the  osmotic  work  done  is  equal  to 


{ 

J  P 


from  which,  by  integration,  we  obtain 


ELECTROCHEMIS  TRY.  4*5 


The  corresponding  electrical  work,  however,  is 
where  TT  is  the  difference  of  potential  between  the  metal 
and  solution  and  SQ  is  the  amount  of  electricity  carried 
by  i  gram  equivalent  of  ions.  We  have  then 


or 


^       P 

*"  log-7' 


from  which  we  see  that  if  P=p,  TT=O. 

£0  for  i  mole  of  a  monovalent  ion  (i.e.,  for  i  equiva- 
lent) is  equal  to  96,540  coulombs.  We  must,  however, 
express  both  sides  of  the  equation  in  electrical  units. 
Since  R=2  cals.,  the  right  side  of  the  equation  must  be 
divided  by  0.2394  in  order  to  obtain  our  result  in  electri- 
cal units.  We  have  then 


2  P 

96,540  XTr(volts)  —  -T  log  — , 

0.4343X0.2394      3  p> 


or 


p 

7r=o.o575  log—  volts  at  17°  C. 
P 


41($  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

For  a  divalent  element  we  must  multiply   s0  by  2; 
hence  if  n  is  the  valence  of  the  metal,  we  have 


p 

\  W«W>^W^    ff^     _  J 

(45)  ~^~1[°%p 

or  for  negative  ions 


0.0002^,      P    0.0002  p 

TT  =  — jf  log  —  = T  log  p  volts. 


If  we  consider  a  cell  composed  of  two  metals,  each 
present  in  a  solution  of  one  of  its  salts,  the  difference 
of  potential  may  arise  (i)  at  the  place  of  contact  of 
the  two  metals,  (2)  at  the  place  of  contact  of  the  two 
solutions,  and  (3)  and  (4)  at  the  points  of  contact  of 
the,  two  electrodes  with  their  solutions.  The  first  can 
be  entirely  neglected  provided  the  temperature  remains 
constant.  The  second,  as  was  mentioned  above,  is 
always  small,  so  that  we  shall  have  to  consider  here  only 
(3)  and  (4)  ,  assuming,  when  necessary,  that  the  value  has 
been  determined  as  on  page  406.  At  17°  the  E.M.F.  of 
the  cell  will  be,  then, 


(46)     ^-*2) 


the   minus   sign   being   used   because   ions   are   formed 
on  the  one  side,  and  consequently  must  disappear  upon 


ELECTROCHEMISTRY  4*  7 

the  other,  for  there  must  always  be  an  equal  amount 
of  positive  and  negative  ions  in  the  solution. 

103.  The  measurement  of  the  potential  difference  be- 
tween a  metal  and  a  solution. — The  method  in  use  for 
the  determination  of  differences  of  potential  is  based 
upon  the  use  of  a  normal  electrode,  i.e.,  an  electrode  in 
a  solution  of  one  of  its  salts,  of  which  the  potential  dif- 
ference is  known.  Making  up  a  cell  by  combining  this 
electrode  with  the  one  to  be  measured,  it  is  possible, 
from  observations  of  the  resulting  electromotive  force, 
to  find  the  value  of  the  unknown  potential  difference. 
The  original  measurement  of  the  potential  difference  of 
the  normal  electrode  was  made  by  forming  a  cell  with 
this  and  another  electrode,  having  a  potential  of  zero 
against  its  solution.  The  electromotive  force  of  this 
combination,  then,  is  identical  with  the  potential  dif- 
ference to  be  found.  Such  an  electrode  with  zero  poten- 
tial is  formed  when  mercury  flows  into  a  solution  of 
an  electrolyte  (KC1),  for  any  differences  of  potential 
existing  between  the  mercury  and  solution  must  eventu- 
ally be  eliminated  by  the  continually  dropping  mercury. 

The  electrodes  in  general  use  are  made  up  according 
to  one  of  the  two  following  schemes: 

HgCl  in  molar  KCl/Hg, 
or 

HgCl  in  o.i  molar  KCl/Hg; 


418 


ELEMENTS  OF  PHYSICAL  CHEMISTRY. 


the  former  giving  a  difference  of  potential  of  —0.56 
volt,  the  latter  of  —0.0613,  both  at  18°.  The  sign  in 
each  case  here  refers  to  the  solution,  i.e.,  the  solution  is 
negative  against  the  metal. 

Fig.  21  shows  the  form  of  the  cell.     The  syphon  tube 
is  so  arranged  that  it  is  always  kept  full  of    the  KC1 


FIG.  21. 

solution.  This  dips  into  the  liquid  of  the  other  electrode, 
provided  they  form  no  precipitate.  In  case  they  do,  a 
molar  solution  of  KNO3  is  used,  into  which  the  siphons 
from  the  two  electrodes  dip.  The  advantage  of  using 
KNO3  as  a  connector  is  that  the  mobilities  of  the  K*  and 
NCV  ions  are  nearly  the  same;  hence  the  passage  of  the 
current  produces  no  change  in  the  concentration  at  the 
two  sides  and  consequently  gives  rise  to  no  difference  of 
potential. 


ELECTROCHEMISTRY.  4*9 

An  example  of  the  use  of  this  electrode  in  measuring 
the  difference  of  potential  of  any  other  is  given  by  the 
following:  For  a  combination  in  the  form 

Zn  -  molar  ZnSO4  -  molar  KC1  HgCl  -  Hg, 

i.e.,  for 

Zn  in  molar  ZnSOi, 

against  the  normal  electrode,  we  find  7r  =  i.o8  volts, 
the  current  going  from  Hg  to  Zn  through  the  wire.  The 
total  E.M.F.  of  the  cell,  then,  is  given  by  the  equation 

RT        PI     RT        P2 

2£Q        &e  pi         £0      ^ep2* 

where  PI  and  pi  refer  to  the  Zn  and  P2  and  p2  to  the  Hg, 
for  that  is  the  only  arrangement  which  would  make  n 
positive.  We  have  then 

RT        Pi 

2£o       e  pi 

~RT  P 

—  log^  —  =  (1.08—0.56)  =0.52  volt; 

i.e.,  Zn  is  negative  against  a  molar  solution  of  its  sulphate 
and  gives  an  E.M.F.  of  0.52  volt.  The  sign  always 
refers  to  the  E.M.F.  of  the  electrolyte  against  the  elec- 


420  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

trade.    Thus,  —  i    means    that    the  solution  is  negative 
and  the  metal  positive,  and  vice  versa. 
In  the  case  of 

Cu-wCuSO4-wKCl  HgCl-Hg, 

we  find  7r=  0.025  volt,  the  current  in  the  wire  going 
from  Cu  to  Hg.  We  have  here,  since  the  current  goes 
from  Hg  to  Cu  in  the  solution, 


Pi 

£o 
or 


RT        Pl 
0.025 —0.56--- log,-, 

RT,      PI 

—  log,— =  -0.025 -0.56  =  -0.585; 


i.e.,  the  electrode  of  Cu  is  positive,  with  a  difference  of 
potential  against  the  electrolyte  equal  to  0.585  volt.  The 
process  which  takes  place  is  shown  more  clearly  by  the 
diagram 


Cu- molar  CuSO4 HgCl  in  molar  KCl-Hg; 

s»-»  0.56     i 
I ^_> , 1 

0.025 


ELECTROCHEMISTRY.  42! 

hence  the  copper  must  be  more  positive  than  the  Hg 
by  0.585  volt., 

104.  The  heat  of  ionization.  —  Knowing  the  difference 
of  potential  existing  between  a  metal  and  a  solution,  and 
its  rate  of  change  with  a  variation  in  the  temperature, 
it  is  possible  from  the  formula  on  page  403  to  find  the 
heat  of  ionization  of  the  metal.  We  found  then  that 


or 

dn 


The  term  Ec,  here,  is  the  heat  produced  when  i  gram 
equivalent  of  the  metal  goes  from  the  metallic  to  the 
ionic  state,  and  its  value  is  given  by  the  formula 


But  (page  405), 

SOT:  =96540X0.2394X7:  volts  =231107:  cals.; 

hence  the  heat  of  ionization  Ec,  for  i  gram  equivalent 
can  be  found  from  the  relation 


=   n- T- 23,110  cals. 


422  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

For  copper  in  copper  acetate  (molar)  at  17°,  7r=o.6, 

.  dn  ..     dn  . 

and  -7^=0.000774,  while  —  for  copper  in  copper  sul- 
phate is  0.000757,  i.e.,  an  average  value  of  —  of 

0.000766  volts,  hence  Ec,  the  heat  of  ionization  of  copper, 
is  8736  cals.  per  gram  equivalent,  or  17,472  cals.  per  mole. 
It  was  in  this  way  that  the  value  for  H  was  determined 
for  use  in  the  table  given  on  page  220. 

105.  Concentration-cells. — a.  In  which  the  electrodes 
have  different  concentrations. — If  the  electrodes  are  made 
of  two  amalgams  of  the  same  metal  in  salts  of  this  metal 
the  E.M.F.  is  given  by  the  formula 


0.0002^        PI     0.0002^.      PZ      . 
-     -Flog-  -T  log-  -volts. 

n  5  n  8    2 


If   the   two   solutions   have   the  same   concentration   of 
metal  ions,  then  p\  =p2>  and  we  have 

0.0002  PI 

TT  =  -  T  log  -77-  volts, 
n  r^2 

when  PI  and  P2  are  the  electrolytic  solution  pressures 
of  the  two  electrodes  with  respect  to  the  dissolved  metal. 
Amalgams  when  dilute  may  be  considered  as  solutions, 
in  which  the  mercury  acts  as  the  solvent;  consequently 
the  osmotic  pressure  of  the  metal  in  the  amalgam  will 


ELECTROCHEMISTRY.  423 

be  proportional  to  the  electrolytic  solution  pressure  of 
the  electrode.  Since  in  all  solutions  the  osmotic  pressure 
is  proportional  to  the  concentration,  we  may  substitute 
for  the  electrolytic  solution  pressures  the  proportional 
terms,  the  concentrations  of  the  metal  in  the  two  elec- 
trodes. We  obtain  then 


0.0002  cl 

TT  =  —     —T  log  —  volts. 
n  6  c2 


This  formula  was  proven  experimentally  by  G.  Meyer 
(Zeit.  f.  phys.  Chem,  7,  477,  1891),  some  of  whose  results 
for  zinc  amalgams  in  a  ZnSC>4  solution  are  given  in  the 

table  below. 

TABLE  XVII. 


T 

Cl 

C2 

T  (obs.) 

*(calc.) 

n°.6 

0.003366 

O.OOOII3O5 

0.0419  v. 

0.0416  v. 

18° 

0.003366 

O.OOOII3O5 

0.0433  v- 

0.0425  v. 

I2°.4 

0.002280 

o  .  0000608 

0.0474  v. 

0.0445  v- 

60° 

0.002280 

0.0000608 

0.0520  v. 

0.0519  v. 

The  results  obtained  by  Meyer  also  proved  the  formula 
to  hold  copper  for  amalgams  in  CuSOi,  etc.,  solutions. 

We  have  assumed  the  metals  to  dissolve  in  mercury  as 
monotomic  molecules  in  this  case,  and  the  assumption 
is  upheld  by  our  results,  not  only  by  this  method  but  also 
by  others. 

If  Zn  had  been  present  in  the  form  of  diatomic  mole- 
cules in  the  mercury,  our  equation  would  have  had  a 
different  value.  For  the  movement  of  the  same  amount 


424  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  ions  as  before,  we  should  have  had,  then,  the  osmotic 
work  equal  to 

1  RT          Cl 

log  — , 

2  0.4343       &C2 

and  since  the  electrical  work  would  have  been  the  same 
as  before,  i.e., 

2X96540?:, 
we  should  have  had 


I     0.0002  Ci 

71  = T  log  —  volts ; 

b 


i.e.,  the  E.M.F.  would  have  been  one-half  what  we  have 
found  it  to  be. 

The  electromotive  force  is  found,  then,  to  be  de- 
pendent only  upon  the  concentration  of  the  metal  in 
the  amalgam  and  its  valence,  and  not  upon  the  nature 
of  the  metal  itself.  The  mercury  has  no  effect  so  long 
as  the  metal  dissolved  in  it  gives  the  larger  E.M.F. 

Another  example  of  the  electrodes  having  different 
electrolytic  solution  pressures,  owing  to  differences  of 
concentration,  is  given  by  cells  of  the  type  of  the  Grove 
gas-battery  in  an  altered  form.  The  electrodes  are  of 
platinized  platinum,  in  which  the  gas  is  absorbed  under 
different  pressures,  and  are  placed  partly  in  the  liquid 


ELECTROCHEMISTRY.  425 

and  partly  in  gas  of  corresponding  (partial)  pressure. 
Such  an  electrode  is  to  be  considered  as  a  perfectly  revers- 
ible gas  electrode,*  i.e.,  one  from  which  ions  of  the 
material  absorbed  as  a  gas  are  given  up,  for  the  metal 
acts  simply  as  a  conductor,  as  has  been  shown  experi- 
mentally by  the  use  of  different  metals,  the  same  result 
being  always  obtained.  In  this  way  reversible  gaseous 
electrodes  of  all  kinds  can  be  made.  Oxygen  as  an 
electrode,  however,  gives  off  ions  of  OH',  since  those  of 
O"  are  not  known  to  exist,  and  absorbs  O  when  the 
OH'  ions  give  up  their  charges  to  it. 

If  we  have  two  electrodes  of  H,  under  different  pres- 
sures, in  contact  with  a  liquid  containing  H*  ions,  we 
shall  obtain  a  certain  E.M.F.  This  may  be  calculated 
in  two  ways,  as  we  did  in  the  case  of  amalgams.  In 
the  second  way,  however,  the  process  is  slightly  different, 
since  one  mole  of  H  gas  forms  two  monovalent  ions 
(p.  146).  The  osmotic  Work  is  equal  to 

RT          PI 
0.4343  CgP2' 

as  before.  The  electrical  work,  however,  which  corre- 
sponds to  this  is  2£o7Tj  for  H2=2H',  hence 

RT 


2^0(0.4343) 


I 

g      ; 


*  The  electrolytic  solution  pressure  here  of  the  gas  electrode  is 
proportional  to  the  nth  root  of  the  gaseous  pressure,  where  n  is  the  num- 
ber of  combining  weights  in  one  formula  weight  of  gas. 


426  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

i.e.,    we   have    2    in    the    denominator   notwithstanding 
that  the  gas   is   monovalent. 

b.  Different  ionic  concentrations. — In  this  type  of 
cell  we  assume  the  electrodes  to  be  of  the  same  metal, 
but  dipping  into  solutions  which  possess  different  ionic 
concentrations  of  the  metal.  An  example  is  given  by 
the  arrangement 

Ag(AgNO3  cone.)  -(AgNO3  dilute) Ag. 
From  the  general  equation  we  have 


RT ,       Pl     RT,       P2 
= log,—         — l°g,:r- 

e  e 


Or,  since  Pi=P2, 

RT 


where  p\  is  the  osmotic  pressure  of  the  Ag'  ions  on  the 
concentrated  side  and  p2  that  on  the  dilute  side. 

From  this  we  see  that  the  E.M.F.  depends  only  upon 
the  ratio  of  the  concentrations,  and  not  at  all  upon  the 
actual  concentrations,  all  of  which  was  found  to  hold 
experimentally  by  Nernst  (1.  c.  p.  406). 

Since   the    osmotic    pressure    is    proportional   to    the 


•^^w-, 

OF  THE      ^    \ 

f    UNIVERSITY   | 
ELECTROCHEMISTR  /i  42  7 

N^^R!j\£X 
concentration,    we    may    substitute    the    latter  "Tor    the 

former;  we  obtain 

RT        Cl 

n= loer  — . 

H£Q        *'  C2 

The  concentration  of  Ag'  ions  in  a  o.oi  molar  solution 
of  AgNO3  is  8.71  times  as  great  as  that  in  a  solution 
which  is  o.ooi  molar  (not  10  times),  hence  the  E.M.F. 
(at  1 8°)  of  a  cell  made  up  of  Ag  in  these  two  concen- 
trations of  AgNO3  should  be 

TT=  0.0002X29 1  log  8.71  =0.054  volt, 
while  0.055  v°to  was  found  by  experiment. 

Since  at  17°  C.  we  have 

°-°575 ,     c\      u 

7T=  —         -log  — VoltS, 

n  c2 

a  concentration  ratio  of  the  ions  equal  to  10,  would  give 
TT =0.0575  volt  for  a  monovalent  ion,  or  0.02875  volt  for 
a  divalent  one. 

This  class  includes  all  cells  made  up  of  solutions 
containing  different  concentrations  of  metal  ions.  In 
case  the  contact  of  the  two  liquids  causes  the  forma- 
tion of  a  precipitate,  another  solution  is  used  between, 
the  two  being  connected  by  small  siphons. 

1 06.  Dissociation  by  aid  of  the  electromotive  force. 
Since  for  concentration-cells  we  have  the  formula 

it  =  • —  -  log  —  volts, 

W£0X  0.4343      °C2 


428  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

or  for  monovalent  metals 


we  can  find  the  ionic  concentration  of  the  metal  on 
the  one  side  provided  that  on  the  other  is  known. 
An  example  of  this  is  the  cell  as  given  by  Morgan: 


Ag- ^ AgN03-KN03- ^KAgCN.-Ag, 


which  gives  an  E.M.F.  of  0.542  volt,  Ag  in  AgNOs  being 
the  positive  pole. 

If,  for  simplicity,  AgNOs  o.i  molar  is  considered 
as  completely  dissociated,  the  concentration  of  Ag'  ions 
will  be  o.i  molar,  and,  since  7  =  273  +  17=290°,  we 
have 


i.e.,   in   the  m/2o  KAgCN2  solution   the   Ag'    ions   are 
present   to   a   concentration   of   3.iXio~n   molar. 

This  method  was  devised  by  Ostwald,  and  is  one  of 
the  most  delicate  known,  for  the  more  dilute  the  one 
solution  is,  i.e.,  the  smaller  the  concentration  of  its  metal 
ions,  the  greater  is  the  E.M.F.  of  the  cell,  and  the  greater 
the  accuracy  of  the  result. 


ELECTROCHEMISTRY.  429 

Another  typical  cell  of  this  kind  is 

Ag  -  ^  AgN03  -  KN03  -  ^  KC1  AgCl  -  Ag, 

which  was  used  by  Goodwin  to  determine  the  solubility 
of  AgCl. 

In  a  saturated  water  solution  of  AgCl  we  may  assume 
without  error  that  the  dissociation  is  complete.  The 
solubility  product  will  then  be 


Since  in  a  pure  water  solution  of  AgCl,  Ag'=Cl', 
the  square  root  of  s,  s  being  determined  under  any 
conditions  at  constant  temperature,  will  give  directly  the 
concentration  of  Ag*  (or  Cl')  in  the  solution.  And,  if 
complete  ionization  may  be  assumed,  this  is  equal  to  the 
solubility  of  AgCl  under  these  conditions. 

For  example,  if  we  find  the  E.M.F.  of  the  cell 

Ag-AgCl  KCl-AgN03-Ag, 

where  ccy  on  the  left  and  c^.  on  the  right  are  known,  we 
can  calculate  c^.  on  the  left,  and  consequently  5  and  the 
solubility  of  AgCl. 

Goodwin  (Zeit.  f.  phys.  Chem.  13,  577,  1894)  found, 


43°  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

when  the  concentration  of  AgNOs  and  KC1  was  o.i 
molar,  an  average  E.M.F.  of  0.45  volt  at  25°.  By  con- 
ductivity measurements  at  25°  we  find 


a  (forAgNO3w/io)=82%  and  a  (for  KClw/io)  =85%, 


•07 

.  J.gJ-1  VX -j"*'/    J-V-y    «->^  J (j      CV1XV4.     V*.      ^iV^A        XX.V^JL"*'/  .J-X/    «-»•< 

so  that 

£l=  0.450 

°  ^2      0.0002  X  298* 

or 


i.e.,  Ag*  ions  are  present  in  a  o.i  molar  KC1  solution 
of  AgCl  to  a  concentration  of  i.94Xio~9  moles  per  liter. 
The  product  s  is  then  found  to  be  equal  to 

1.94  X  io~9  Xo.o85  =s  =  1.64  X  io~10 
and 


i.e.,  AgCl  in  a  saturated  water  solution  at  25°  is  present 
to  a  concentration  of  1,28  Xio~5  moles  per  liter. 

A  cell  of  this  sort  may  be  arranged  with  bromides, 
iodides,  etc.,  in  place  of  the  chlorides,  which  was  also 
done  by  Goodwin,  and  the  solubility  of  AgBr  and  Agl 
determined. 

Another  illlustration  of  this  method  of  determining 
dissociation  is  furnished  by  Ostwald's  arrangement  for 


ELECTROCHEMISTRY.  43  ' 

determining  the  dissociation  constant  for  water.  The 
scheme  is  a  gas-cell  in  the  form 

H-Acid  ........  Base-H, 

when  the  two  H  electrodes  have  the  same  solution  pres- 
sure. The  E.M.F.  of  such  a  cell,  deducting  the  E.M.F. 
caused  by  the  contact  of  the  two  solutions,  is,  for  molar 
solutions  at  17°  C., 

TT  =0.0575  log  A 

Here,  contrary  to  the  case  for  electrodes  of  different 
gaseous  concentration,  the  2  does  not  appear  in  the 
denominator  (page  425),  for  that  factor  simply  refers 
to  the  process  taking  place  between  the  electrode  and 
the  gas.  In  this  case  the  electrodes  remain  constant 
and  the  same,  and  only  the  osmotic  pressures  of  the 
H'  ions  is  to  be  considered.  By  experiment  7r=o.8i 
volt;  hence,  since  £1=0.8  (i.e.,  the  normal  acid  is  80% 
dissociated), 


or 

c2=o.8Xio~14. 

Since  this  is  the  concentration  (in  terms   molar)  of  the 
H*  ions  in  the  base,  and  the  concentration  of  OH'  ions 


432  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

in  contact  with  them  is  0.8,  the  ionic  product  for 
H20  is 

o.8Xo.8Xio~14. 

Since  H2O  dissociates  into  H*  and  OH'  ions,  the  con- 
centration of  H'  and  OH'  ions  in  pure  water  is  the  same, 
and  is  equal  to 


V  (o.8)2Xio-14  =0.8 X  io~7  =H*  and  OH'  ions  in  H2O. 

From  the  conductivity  also,  at  this  temperature, 
Kohlrausch  found 

o.8Xio~7. 

It  would  be  possible  in  this  case  to  use  oxygen  elec- 
trodes, by  which  the  same  final  result  would  be  obtained, 
but  platinum  black  absorbs  very  little  oxygen,  so  that 
the  results  are  not  as  certain  as  with  hydrogen. 

For  the  measurement  of  the  concentration  of  ions  of 
mercury  the  normal  electrode  serves  as  the  basis.  The 
concentration  of  Hg2  ions  in  the  normal  electrode  has 
been  found  to  be  3.5Xio~18  moles  per  liter  (Ley  and 
Heimbucher,  Zeit.  f.  Electrochem.,  10,  303,  1904).  Since 

Hga 

•  -  =120,  in  the  presence  of  metallic  mercury,  accord- 
ing to  Abel  (Zeit.  f.  anorg.  Chem.,  26,  377,  1901),  it  is 
possible  to  find  the  concentration  of  Hg"  in  the  normal 


ELECTROCHEMISTR  Y.  433 

electrode;  it  is  equal  to  2.9Xio~20  moles  per  liter.  By 
measurements  of  the  electromotive  force  of  a  combination 
of  mercury  in  a  solution  containing  Hg"  or  Hgs  ions, 
with  the  normal  electrode  we  can  thus  determine  the  con- 
centration of  these  ions.  For  further  details  as  to  this,  see 
Sherrill  and  Skowrouski,  J.  Am.  Chem.  Soc.,  27,  30,  1905. 
107.  Electrolytic  solution  pressure. — It  is  possible  from 
the  potential  difference  between  a  metal  and  its  salt 
solution  to  calculate  the  solution  pressure  in  atmos- 
pheres. The  equation  is 


or  at  if 


If  we  use  normal  solutions,  p  is  equal  to  22. 
atmospheres  if  completely  dissociated.  The  pressure  of 
the  ions,  in  case  the  salt  is  not  completely  dissociated, 
is  then  easily  found  (i.e.,  a X 22. 4}^- J  atmospheres).  Since 
TT  can  be  determined,  the  value  for  P  can  be  found  for 
all  metals.  In  the  table  on  page  434  the  values  in 
atmospheres,  as  determined  by  Neumann,  are  given: 

These  values  it  must  be  remembered  are  merely  sym- 
bolical, for  the  gas  laws  may  not  be  applied  to  such  an 


434  ELEMENTS  OP  PHYSICAL   CHEMISTRY. 

extent.  The  relation  between  these  numbers,  however, 
are  the  experimentally  determined  relative  effects  of  the 
metals,  as  a  glance  at  the  formula  will  show. 


TABLE  XVIII. 

SOLUTION  PRESSURES  OF  METALS. 

Zinc 9 . 0X io18  atmospheres 

Cadmium 2 .  7X10* 

Thallium 7.7Xio2 

Iron 1.2X10* 

Cobalt 1.9X10° 

Nickel 1.3X10° 

Lead i.iX io~3 

Hydrogen 9.9X10"* 

Copper 4.8Xio~20 

Mercury i.iX io~ic 

Silver 2.3Xio~17 

Palladium i.Xio-30 


The  influence  of  the  dilution  on  the  difference  of 
potential  depends  upon  the  magnitude  of  the  electrolytic 
solution  pressure.  If  the  latter  is  great,  dilute  solutions 
are  necessary  for  maximum  action;  if  small,  more  con- 
centrated solutions  are  to  be  preferred. 

108.  Cells  with  inert  electrodes. — Elements  of  this 
type  'have  electrodes  in  solutions  which  do  not  contain 
the  ions  of  the  electrodes.  Such  cells  are  also  known  as 
oxidation  and  reduction  cells.  An  example  is  given  by 
the  arrangement 

plat.Pt-FeC!3sol SnQ2sol.  -plat.Pt. 


ELECTROCHEMISTR  Y.  435 

The  Sn"  ions  go  over  into  Sn::  ions,  taking  up  the  elec- 
tricity set  free  by  the  transformation  of  Fe"*  ions  into 
those  of  Fe".  The  Fe  side,  then,  is  positive  and  the 
Sn  side  negative. 

The  current  in  these  cells  is  due  to  the  oxidation 
on  the  one  side  and  the  reduction  on  the  other.  An 
oxidizing  substance  (i.e.,  one  which  is  reduced)  is 
any  substance  from  which  negative  ions  are  formed  or 
upon  which  positive  ions  give  up  their  charges.  A 
reducing  substance  is  one  in  which  negative  ions  are 
given  up  or  positive  ions  formed. 

109.  Processes  taking  place  in  the  cells  in  common 
use. — All  cells  give  an  E.M.F.  which  is  equal  to  the 
difference  of  the  differences  of  potential  at  the  two  single 
electrodes;  but  in  many  cases  the  action  is  complicated 
by  other  processes.  In  the  following  the  effect  of 
depolarizers  is  given,  as  well  as  the  change  in  the  cell 
during  action. 

Clark  cell. — This  cell,  as  already  mentioned,  is  used 
as  a  standard  of  E.M.F.     The  scheme  is 
Hg-Hg2SO4-ZnSO4-Zn. 

The  Hg2SC>4  is  not  quite  insoluble,  therefore  a  small 
concentration  of  Hg  ions  does  exist  in  the  solution.  The 
Zn"  ions  with  their  high  solution  pressure  are  driven 
from  the  electrode  into  the  ZnSC>4  solution,  and  force 
before  them  to  the  electrode  the  ions  of  H',  since  the 


436  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

solution  can  contain  only  the  same  concentration  of  posi- 
tive as  negative  ions.  The  H*  ions  give  up  their  charges 
to  the  electrode,  which  becomes  positive,  the  Zn  being 
negative  from  loss  of  positive  ions.  The  current  goes, 
then,  externally  through  the  wire  from  the  Hg  to  the  Zn, 
and  internally  from1  the  Zn  to  the  Hg.  If  the  resistance 
placed  against  the  cell  is  great  enough,  only  an  infini- 
tesimal amount  of  Hg2  ions  will  be  driven  from  the  solu- 
tion, and  the  cell  remains  constant.  If  the  current  is 
passed  through  a  small  resistance,  however,  all  the  Hg2' 
ions  are  soon  precipitated  upon  the  electrode  and  the 
E.M.F.  is  reduced,  i.e.,  the  cell  is  polarized.  If  allowed 
to  stand,  however,  more  Hg2SO4  will  -dissolve  and  the 
original  E.M.F.  is  attained. 

The  Leclanche  cell  consists  of  a  solution  of  ammonium 
chloride,  in  which  we  have  two  electrodes,  Zn  and 
C+MnO2.  The  action  of  the  MnO2  is  to  prevent 
polarization,  the  processes  taking  place  without  it  and 
with  it  being  as  follows: 

In  the  cases  without  MnO2  the  Zn  with  its  high  solu- 
tion pressure  goes  into  solution,  driving  before  it  the 
other  positive  ions,  i.e.,  those  of  NH4-  These  ions  de- 
compose on  losing  their  changes,  forming  NH3  and  H 
gases.  The  bubbles  of  H  collect  upon  the  electrode  and 
are  absorbed  and  so  given  off  to  the  air,  but  as  this  process 
is  slow,  other  ions  are  prevented  from  giving  up  their 
charges  and  consequently  the  E.M.F.  decreases.  It  is  to 


ELECTROCHEMISTRY.  437 

get  rid  of  this  action  of  the  H  gas  that  the  MnO2  is  used. 
In  contact  with  water  we  have,  then,  to  a  small  extent* 
a  solution  of  MnO2,  which  dissociates  according  to  thg 
scheme 


MnO2  4-  aHaO  =  Mn:  : 


These  tetravalent  ions  of  Mn  have  the  tendency  to  go 
into  the  divalent  state  by  giving  up  two  equivalents  of 
electricity,  i.e.,  to  form  the  ions  Mn".  In  consequence 
of  this  the  Mn::  ions  with  those  of  NHi  are  driven  to 
the  electrode  by  the  Zn"  ions,  and,  since  they  give  up  two 
equivalents  of  electricity  more  readily  than  any  other 
ion  gives  up  its  entire  charge,  the  electricity  is  given  up 
without  any  substance  which  might  cause  polarization. 
We  have,  then,  MnCl2  (i.e.,  Mn"  ions)  formed  in  the 
solution.  The  process  continues  so  long  as  there  is  solid 
Zn  or  MnO2  present. 

Bichromate  cell.  —  This   cell  is  arranged  according  to 
the  scheme 


The   action    of  the   two   substances   in    solution   forms 
bichromic  acid  (H2Cr2O7).     This  dissociates  into 

2H'  +  Cr207" 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 
to  a  large  extent,  and  to  a  smaller  degree  as  follows : 

H2Cr2O7  +  sH2O  =  2Cr : :  + 1 2  •  OH'. 

These  hexavalent  Cr  ions  have  the  tendency  to  give 
up  three  equivalents  of  electricity  and  to  go  into  the 
trivalent  state  (i.e.,  Cr").  Accordingly  the  Zn"  ions 
which  are  forced  from  the  electrode  drive  before  them 
the  Cr::  ions,  which  give  up  three  of  their  equivalents 
and  become  Cr",  remaining  as  such  in  the  solution  as 
ions  in  equilibrium  with  SO^'  ions  (i.e.,  as  Cr2(SO4)s). 
Finally,  then,  we  have  a  solution  of  Cr2(SO4)3  left  in  the 
jar.  This  change  in  the  number  of  electrical  equiva- 
lents by  a  change  of  valence  always  takes  place  more 
readily  than  the  change  from  the  ionic  to  the  elemental 
state  and  is  of  great  value  as  a  means  of  preventing 
polarization. 

Accumulators. — The  action  of  the  lead  accumulator 
or  storage-cell  also  depends  upon  a  change  of  valence. 
Any  reversible  cell  can  be  recharged  after  it  is  used  up 
by  the  passage  of  a  current  through  it  in  the  direction 
opposite  to  that  in  which  it  goes  of  itself.  The  lead  cell, 
however,  is  generally  used  for  the  purpose  owing  to  its 
high  E.M.F.  Before  charging  it  consists  of  two  ,lead 
plates,  one  of  which  is  coated  with  litharge  (PbO)  in 
a  20%  solution  of  sulphuric  acid.  If  the  current  is 
passed  through  these  plates  (the  PbO  being  positive), 


ELECTROCHEMISTRY.  439 

the  PbO  is  transformed  into  PbO2,  lead  superoxide  (or 
supersulphate),  while  spongy  lead  is  deposited  upon  the 
other  electrode.  The  flow  of  current  is  now  stopped, 
and  the  cell  is  charged. 

The  PbO 2  is  soluble  to  a  small  degree  and  ionizes  as 
follows : 

Pb02  +  2H20=Pb::  +4OH'. 

These  tetravalent  Pb  ions  have  the  tendency  to  give  up 
two  of  their  electrical  equivalents  and  to  go  into  the 
divalent  form.  Since  this  is  true,  the  Pb  electrode 
must  have  the  higher  solution  pressure,  and  the  ions 
formed  from  it  will  drive  those  of  Pb::  to  the  electrode, 
where  they  will  lose  two  charges  of  electricity  and  become 
divalent.  This  will  continue  as  long  as  PbO 2  is  present, 
i.e.,  until  it  is  all  transformed  into  the  divalent  state, 
PbSO4.  Other  theories,  by  Liebenow  and  others,  have 
also  been  advanced  to  explain  this  cell,  but  the  reader 
must  be  referred  elsewhere  for  them  (see  Dolezalek  and 
von  Ende,  The  Theory  of  the  Lead  Accumulator). 

Dolezalek  has  been  able  to  calculate  the  value  of 
such  a  cell  from  the  concentration  of  acid  and  the  vapor- 
pressure  of  the  solution,  and  finds  an  excellent  agree- 
ment between  theory  and  experiment.  This  proves  the 
process,  according  to  him,  to  be  a  primary  one  such  as 
the  theory  of  Le  Blanc  and  that  of  Liebenow  would 
make  it. 


44°  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 


D.  ELECTROLYSIS  AND  POLARIZATION. 

no.  Decomposition  values. — We  must  now  consider 
the  processes  which  take  place  when  a  current  is  passed 
through  a  solution  from  electrodes  which  are  not  affected 
by  the  liquid,  i.e.,  inert  ones,  as  platinum,  gold,  carbon, 
etc.  The  current  causes  the  deposition  of  the  positive 
and  negative  ions  on  the  electrodes.  If  the  primary 
current  is  disconnected  at  any  time  and  the  poles  of  the 
decomposition-cell  connected  with  a  galvanometer,  we 
observe  a  current  in  the  opposite  direction  which  becomes 
weaker  and  weaker,  until  it  finally  disappears  entirely. 
This  is  called  the  current  of  polarization,  and  its  E.M.F. 
is  the  E.M.F.  of  polarization.  This  is  caused  by  the 
tendency  of  the  substance  precipitated  upon  the  elec- 
trode to  go  back  into  solution  in  the  dissociated  form. 

It  has  been  found  that  a  certain  minimum  of  E.M.F. 
is  necessary  to  cause  the  steady  electrolysis  of  any  solu- 
tion. If  the  E.M.F.  used  is  smaller  than  this,  the  cur- 
rent goes  through  for  an  instant  and  then  ceases;  but 
at  this  point  or  above  it  the  process  goes  steadily.  This 
was  found  by  Le  Blanc,  who  determined  the  minimum 
values  for  a  large  number  of  liquids  and  solutions.  The 
table  below  gives  these  values  for  molar  solutions  of  the 
salts  which  separate  metals. 


ELECTROCHEMISTR  Y. 


441 


TABLE  XIX. 

DECOMPOSITION    VALUES    FOR   SALTS. 


ZnSO4  =  2.35  v. 
Cd(N03)2=i.98v. 
ZnBr3  =  i .  80  v. 
NiSO4  =  2.09V. 
NiCl,  =i.8sv. 
Pb(N03)2=i.52v. 


Ag(NO)4=  0.70V. 
CdSO2  =2. 03v. 
CdCl2  =i.88v. 
CoSO4  =  i .  94  v. 
CoCl,  =  i .  78  v. 


The  values  for  Cd(NO3)2  and  CdSO4  show  that  those 
for  the  nitrates  and  sulphates  of  the  same  metal  are 
nearly  the  same. 

The  following  tables  contain  the  values  for  acids 
and  bases.  Here  there  is  a  certain  maximum,  which 
is  reached  by  many  and  exceeded  by  none,  which  is 
about  1.70  volts. 

TABLE  XX. 

DECOMPOSITION   VALUES   FOR   THE    ACIDS. 

Dextrotartaric  =  i .  62  v. 

Pyrotartaric  =  i .  5  7  v. 

Trichloracetic  =  i .  5 1  v. 

Hydrochloric  =  i .  31  v. 

Oxalic  =o.95v. 

Hydrobromic  =  o .  94  v. 

Hydriodic  =  o .  5  2  v. 

TABLE  XXI. 

DECOMPOSITION  VALUES  FOR  THE   BASES. 


Sulphuric               = 

.67  v. 

Nitric                      = 

.69  v. 

Phosphoric             = 

.  7ov. 

Monochloracetic  = 

.72V. 

Dichloracetic         = 

.66v. 

Malonic                  = 

[.72  v. 

Perchloric              =  ] 

[  .  65  v. 

Sodium  hydroxide 
Potassium  hydroxide     = 
Ammonium  hydroxide  = 
Methylamine  n/4 
n/2 


.  69  v.  n/4 
.67  v.  n/2 

-74v.  «/8 

•75 
.68 

•74 


442  ELEMENTS  OF  PHYSICAL    CHEMISTRY. 

The  alkalies  and  alkaline  earths  when  combined  with 
the  largely  dissociated  acids,  i.e.,  those  with  decomposi- 
tion values  of  about  1.70  volts,  show  approximately  the 
same  value,  viz.,  2.20  volts.  The  chlorides,  bromides, 
and  iodides  have  lower  values,  which  are  independent 
of  the  alkali  metal. 

It  will  be  observed  that  all  the  acids  and  bases  with 
the  maximum  value  give  off  hydrogen  and  oxygen  upon 
electrolysis.  Those  with  the  lower  values,  which  in 
more  dilute  solutions  give  off  oxygen  and  hydrogen, 
also  reach  this  maximum  value.  Thus  for  HC1  at 
different  dilutions  we  have: 

DECOMPOSITION  VALUES  FOR  HC1. 


Concentration.       n 
2  normal  =1.26 
1/2    "     =i-34 
1/6    "     =1.41 


Concentration.          n 
i/ 1 6  normal  =1.62 
1/32      »       =1.69 


At  the  dilution  n/^2  H  and  O  are  given  off  instead 
of  H  and  Cl,  as  at  the  strength  of  2  normal,  and  the 
maximum  is  reached. 

All  the  above  results  are  -for  platinum  electrodes. 
For  gold  the  values  are  slightly  different,  but  the  rela- 
tion remains  the  same. 

Hi.  Theory  of  polarization. — The  process  taking 
place  upon  an  electrode  during  electrolysis  has  been 
studied  by  Le  Blanc,  who  has  been  able  to  make  the 
action  quite  clear.  He  measured  the  difference  of 


ELECTROCHEMIS  TR  Y.  44  3 

potential  between  the  cathode  and  its  solution  after  the 
passage  of  the  current,  varying  the  E.M.F.  from  o  up  to 
the  decomposition  value  of  the  solution.  At  the  de- 
composition value  he  found  the  potential  difference  to 
be  the  same  as  that  exhibited  by  the  solution  in  which 
is  placed  a  stick  of  the  metal.  Thus  a  molar  solution  cf 
CdSC>4  is  decomposed  steadily  at  2.03  volts,  at  which 
the  difference  of  potential  between  the  cathode  and 
solution  is  found  to  be  +0.16  volt.  A  stick  of  cad- 
mium in  a  normal  solution  of  CdSC>4  gives  an  E.M.F. 
equal  to  -f  0.16  volt,  i.e.,  the  metal  is  negative.  For 
some  solutions  this  difference  of  potential  between 
electrode  and  solution  is  found  to  be  given  without 
the  maximum  being  reached.  This  is  true'for  AgNO3, 
which  decomposes  at  0.70  volt.  The  reason  for  this 
is  the  negative  solution  pressure  of  the  Ag,  which  causes 
ions  to  be  deposited  upon  the  metal,  even  without  the 
current. 

When  an  indifferent  electrode  is  placed  in  a  salt  solu- 
tion, a  small  amount  of  ions  must  precipitate  upon  it; 
otherwise,  from  the  equation 


TT  will  be  infinite,  since  P=o;  and  if  that  is  true  it  would 
be  possible  to  arrange  a  perpetual  motion.     Metal  ions 


444  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

will  then  separate  upon  the  electrode  until  the  tendency 
for  them  to  go  into  solution  in  the  ionic  form  just  com- 
pensates the  separating  force.  In  this  way  the  electrode 
will  be  charged  positively,  and  the  solution  negatively, 
so  that  there  will  be  between  them  a  certain  difference 
of  potential.  The  value  of  this  difference  will  depend 
upon  the  amount  of  metal  upon  the  electrode,  and  will 
not  necessarily  be  as  large  as  that  for  the  massive  metal 
(compare  with  concentration  of  H  in  a  platinum-black 
electrode).  If  we  now  connect  the  cell  with  a  primary 
cell  at  a  low  E.M.F.,  more  metal  will  be  deposited, 
but  this  will  increase  P,  so  that  no  more  deposition  will 
take  place  at  that  E.M.F.  If  the  potential  is  then 
increased,  P  will  again  increase  and  further  deposition 
will  be  prevented.  Finally,  when  the  E.M.F.  used  is 
such  that  P  receives  its  maximum  value,  i.e.,  that  which 
the  massive  metal  possesses,  then  the  deposition  will 
take  place  steadily.  Here  we  assume  the  osmotic  pres- 
sure of  the  metal  ions  to  remain  constant.  For  strong 
currents  this  is  not  true;  it  decreases,  and  hence  the 
deposition  becomes  more  and  more  difficult  and  the 
potential  difference  at  the  cathode  increases.  It  is  not 
difficult,  however,  to  keep  the  value  constant  for  a  short 
time,  and  thus  to  find  the  difference  of  potential  at  the 
cathode. 

At  the  same  time  that  this  takes  place  negative  ions 
are  separated  upon  the  anode  and  the  process  is  exactly 


ELECTROCHEMISTRY.  445 

analogous.  If  oxygen  is  the  gas  which  separates,  its 
concentration  increases  until  the  maximum  is  reached 
(=P)  and  the  gas  is  given  off  in  the  air. 

From  the  above  it  is  possible  to  understand  why  a 
certain  minimum  E.M.F.  is  required  to  cause  a  steady 
electrolysis.  The  separation  takes  place  only  when 
the  concentration  of  ions  around  the  electrodes  reaches 
a  maximum,  i.e.,  not  until  the  osmotic  pressure  exerted 
by  them  is  great  enough  to  overcome  the  solution  pres- 
sure of  the  metal.  From  the  case  of  AgNO3  it  would 
seem  that  this  maximum  of  concentration  need  not 
be  reached  on  both  sides  at  the  same  time.  The  nature 
of  the  electrodes,  so  long  as  they  are  of  an  inert  sub- 
stance, has  no  influence  except  for  gases,  and  then  in 
the  following  way:  The  cell  platinized  Pt+H  gas 
—  H2SO4  — O  gas  -f  platinized  Pt  gives  an  E.M.F.  of 
1.07  volts.  If  1.07  volts  are  passed  through  the  cell 
in  the  opposite  direction,  then  we  shall  have  equilibrium. 
If  smaller  than  this,  H2O  is  formed  at  the  electrodes 
from  the  gas  in  the  electrode  and  the  ions  in  the  solu* 
tion.  If  larger  than  1.07,  H  and  O  are  given  off  from 
the  electrodes.  We  have  here  the  gas-cell  which  gives 
the  value  1.07  decomposing  water,  and  reversing  at 
i. 08.  When  polished  platinum  electrodes  are  used  for 
the  decomposition,  however,  1.68  volts  are  necessary  for 
steady  electrolysis,  the  difference  between  the  1.68  and 
the  1.07  being  0.6  volt.  The  difference  in  these  two 


446  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

values  is  assumed  to  be  due  to  the  different  ion  which 
is  separated  in  the  two  cases.  Water  may  be  considered 
as  dissociated  as  a  dibasic  acid: 


OH'=H' 


At  the  value  1.08  volts  the  O"  ions  present  separate, 
causing  water  to  decompose  for  a  time  and  then  as 
they  are  used  up  the  process  ceases.  The  1.07  volts 
of  the  gas-cell  are  due  to  this  ion  O",  for  the  platinum 
black  gives  it  up  to  the  solution  in  that  form  after  having 
absorbed  oxygen  gas.  When  OH'  ions  separate  they 
unite  to  form  H^O  and  O, 

4OH'=2H2O+O2, 

and  the  force  necessary  to  do  this  is  1.68  volts.  Where 
the  concentration  of  OH'  ions  is  great,  in  bases  for  ex- 
ample, the  O"  ions  would  be  present  to  a  greater  extent 
than  in  acids,  and  the  decomposition  at  1.08  should  be 
more  marked  than  with  acids,  as  it  is  found  to  be. 
HC1  has  too  few  OH'  and  O"  ions  to  carry  any  amount 
of  current,  so  that  in  strong  solutions  Cl  separates  at 
1.31  volts,  at  which  point  water  cannot  be  decomposed 
into  hydrogen  and  oxygen. 


ELECTROCHEMISTRY.  447 

If  we  have  in  a  decomposition  vessel  one  large  platin- 
ized electrode  and  one  very  small  pointed  one  and  use 
the  latter  as  cathode,  the  large  one  as  anode,  we  get 
a  rapid  decomposition  of  water  at  i.i  volts,  H  being 
given  off  at  the  point,  the  oxygen  being  absorbed  in 
the  platinum  black.  We  get  then  a  reversal  of  the  gas- 
battery.  Using  the  point  as  anode,  we  get  bubbles 
of  oxygen  first  at  1.68  volts,  H  being  absorbed  by  the 
large  electrode.  An  example  of  the  use  of  such  a 
process  is  the  Hildburgh  cell  (J.  Am.  Chem.  Soc.,  22, 
300,  1900),  which  rectifies  an  alternating  current.  If  the 
value  of  the  alternating  current  is  below  1.68,  current 
can  go  through  when  the  point  is  the  cathode  and  is 
stopped  when  this  is  the  anode,  as  it  is  by  the  next  reversal; 
so  that  we  get  from  this  a  current  made  up  of  a  series 
of  impulses  all  in  the  same  direction  instead  of  the 
alternating  current  which  entered.  A  current  of  large 
voltage  can  then  be  rectified  by  passage  through  a  series 
of  such  cells. 

112.  Primary  decomposition  of  water. — The  E.M.F.  of 
decomposition  of  a  cell  giving  off  hydrogen  and  oxygen 
is  dependent  upon  the  concentration  of  the  two  ions  H* 
and  OH',  but  independent  of  the  nature  of  the  electroltye. 
The  E.M.F.  is  the  same  for  acids  and  bases  so  long  as 
only  H  and  O  are  separated.  Since  by  the  law  of  mass 
action  the  product  of  H*  and  OH'  ions  in  a  solution 
must  always  be  the  same,  it  follows  that  for  all  electro- 


448  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

lytes,  since  the  E.M.F.  of  the  cell  must  be  the  sum  of 
those  af  the  two  electrodes,  the  minimum  value  must 
be  the  same  for  all  substances  which  give  off  H  and  O. 
With  the  exception,  then,  of  solutions  of  metallic  salts, 
which  are  decomposed  by  H,  and  the  chlorides,  bromides, 
nnd  iodides,  which  are  decomposed  by  O,  the  ions  of 
water  only  are  the  factors  in  the  decomposition  of  solu- 
tions, and  not  those  of  the  dissolved  salt.  Excluding 
those  solutions  mentioned  above,  then,  all  solutions 
when  electrolyzed  show  primary  decomposition  of  water. 
The  current  is  conducted  by  all  the  ions  in  the  solu- 
tion. At  the  electrodes,  however,  that  process  takes 
place  which  requires  the  smallest  expenditure  of  work, 
and  that  is  the  separation  of  H  and  O  as  gases.  In  the 
case  of  the  electrolysis  of  K2SO4  in  solution,  when  the 
current  is  not  too  strong,  there  is  no  necessity  for  assum- 
ing that  the  K'  and  SO'4'  ions  are  separated  upon  the  elec- 
trode and  then  redissolved.  They  simply  collect  around 
the  electrode,  but,  as  the  H  and  O  separate  more  easily, 
these  are  forced  out.  With  a  stronger  current,  of  course, 
it  is  possible  to  cause  the  separation  of  K  and  SC>4,  for 
then  the  H*  and  OH'  ions,  being  present  to  but  a  small 
extent,  cannot  carry  all  the  current,  and  the  work  of 
separation  of  the  ions  of  the  salt  is  smaller  than  that 
necessary  to  remove  the  very  much  diminished  amount 
of  OH'  and  H'.  The  formation  and  decomposition  of 
H2O  are  reversible  processes,  so  that  there  is  no  loss 


ELECTROCHEMISTRY  449 

of  work  connected  with  them,  while  with  the  secondary 
action  there  would  be.  The  H*  and  OH'  ions  are  formed 
as  they  are  used  up,  i.e.,  more  water  dissociates. 

All  bases  and  acids  must  have  the  same  E.M.F.,  for 
the  product  of  the  H*  and  OH'  ions  is  always  the  same. 
For  salts  we  should  obtain  a  higher  value,  since  upon 
the  electrode  at  which  H  separates  a  base  is  formed,  and 
OH'  ions  increase  in  number,  driving  back  those  of 
H',  so  that  the  difference  of  potential  on  that  electrode 
increases,  i.e.,  P  remains  the  same,  while  p  decreases.  On 
the  other  electrode  H'  ions  collect  and  exert  the  same 
decreasing  effect  upon  the  OH'  ions.  The  smaller  the 
dissociation  of  the  base  and  acid  formed  the  smaller 
naturally  this  influence  will  be. 

In  the  case  of  HC1  (w/i),  Cl  gas  is  given  off  steadily 
at  a  smaller  E.M.F.  than  for  the  oxygen  acids.  As  the 
acid  becomes  more  dilute,  however,  the  amount  of  H" 
and  Cl'  ions  decreases,  until  finally  there  is  such  a  large 
concentration  of  OH'  ions  present,  as  compared  with  that 
of  Cl',  that  O  separates  the  more  readily.  This  explains 
the  results  already  given  for  HC1  in  different  dilutions. 

113.  Electrolytic  separation  of  metals  by  graded  elec- 
tromotive forces. — As  we  have  seen,  the  salts  of  the 
different  metals  have  different  decomposition  values. 
From  this  fact  Freudenberg*  showed  how  it  is  possible 
to  separate  metals  quantitatively.  It  is  only  necessary 

*  Zeit.  f.  phys.  Chem.,  12,  97,  1893. 


45°  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

to  find  two  salts,  i.e.,  one  of  each  metal,  which  have 
decomposition  values  as  far  apart  as  possible.  If  now 
a  certain  E.M.F.,  lying  between  these  limits,  is  passed 
through  the  cell,  the  metal  with  the  lower  decomposition 
value  will  separate;  after  that  is  separated,  the  current 
will  cease  and  it  is  only  necessary  to  raise  the  E.M.F. 
in  order  to  deposit  the  other.  As  the  concentration  of 
the  ions  in  the  salt  of  the  metal  separated  first  decreases, 
it  is  necessary  to  raise  the  E.M.F.  slightly.  This  amount 
is  small  and  may  be  readily  calculated  from 


RT 


Thus  if  p  decreases  from  o.i  normal  to  o.oooooi  normal 
(the  limit  by  analytical  means),  x  must  be  increased 
only  0.3  volt  for  a  monovalent  element  and  but  half 
that  amount  for  a  divalent  one. 

An  example  of  this  is  given  in  the  table  below. 
AgNO3=o.7o  and  Pb(NO3)2  =  i.52.  With  the  two 
solutions  present  the  Ag  will  be  entirely  deposited  by 
an  E.M.F.  of  less  than  one  volt;  after  this  is  done  the 
E.M.F.  may  be  raised  to  1.52  or  more  and  all  the  lead 
deposited.* 

In  this  way  it  is  possible  to  make  separations  which 
cannot  be  made  by  varying  current  strength. 

*  The  student  is  referred  for  further  information  to  the  original 
article. 


ELECTROCHEMISTRY.  451 

The  following  table  gives  the  separation  value  of  a 
few  ions.  Here  they  are  based  on  the  value  of  H*  taken 
as  zero.  Thus  if  the  value  for  H'  is  added  to  that  of 
OH',  we  have  the  decomposition  voltage  of  H2O  1.68. 
The  values  are  for  molar  concentration. 


Ag- 

=  -0.78 

r      =0.52 

Cu' 

=  -o-34 

Br'       =0.94 

H- 

=  4-Q.o 

O"      =  i.  08  (in  acid) 

Pb' 

=  +0.17 

Cl'       =1.31 

Cd' 

=  +0.38 

OH'    =1.68  (in  acid) 

Zn' 

=  4-0.74 

OH'    =0.88  (in  base) 

SO/'  =1.9 

HSO/  =  2.6 

The  values  of  O"  and  OH'  are  true  in  the  presence  of  a 
normal  solution  of  H*  ions.  If  we  have  H*  and  OH' 
in  a  base,  the  above  value  of  H*  becomes  0.8  and  the 
value  of  OH'  and  O"  is  decreased  by  0.8. 

Speketer  (Zeit.  f.  Elektrochem.,  4,  539,  1898)  has 
applied  the  table  above  to  the  quantitative  determina- 
tion of  Cl,  Br,  and  I  in  solutions.  A  platinum  cathode 
and  a  silver  anode  are  placed  in  a  normal  solution  of 
H2SO4  containing  the  mixed  halides.  The  minimum 
electromotive  force  of  decomposition  of  this  system,  i.e., 
solution  of  Ag  and  separation  of  H,  is  given  by  the  equa- 
tion 

i      Pl  i      P2 

7T  =0.0575  1(>g  ^  -0-0575  lQg  fr 

at  17°,  where  P\  and  P^  are  the  electrolytic  solution 
pressures  of  the  Ag  and  H,  pi  and  p2  being  the  osmotic 


45 2  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

pressures  of  the  ions  Ag*  and  H'.  We  have  then,  taking 
the  decomposition  value  of  H*  as  zero  and  that  of  Ag'  as 
—  0.78  (since,  when  pi=i,  0.0575  log  P\  =  —0.78), 

P1 

*  =°-°575  log  —  =0.0575  log  pi  -»°-°575  log  Pi 

=  -0.78-0.0575  log^i. 

pij  however,  the  concentration  of  Ag'  ions,  can  only 
reach  a  certain  value  in  the  presence  of  Cl,  Br,  and  I, 
which  value  is  regulated  by  the  solubility  products  of 
AgCl,  AgBr,  and  Agl.  The  solubilities  at  25°  are 
respectively  i.25Xio~5,  6.6Xio~7,  and  0.97  Xio~8.  If 
I  is  present  in  the  ionic  state  to  the  concentration  o.i  mole 

per   liter,    A1=L_^Z L}  from  which  7r1  =  +o.ooF, 

o.i 

i.e.,  the  system  acts  as  a  cell  and  gives  a  current,  Ag 
dissolving  and  H  being  evolved.  For  Br,  x=—  o.i2F; 
and  for  Cl,  TTJ=— o.2F.  If  I'  ions  are  reduced  in  con- 
centration to  o.oooi  mole  per  liter,  n  becomes— 0.084. 
In  this  way,  by  varying  the  E.M.F.,  the  iodide  and 
then  the  bromide  are  transformed  into  the  silver  salt, 
and  the  difference  in  weight  of  the  anode  before  and 
after  the  experiment  gives  the  weight  of  the  one  which 
has  been  separated.  Cl  is  usually  determined  by  titra- 
tion  after  the  Br  and  I  have  been  removed. 


CHAPTER  X. 
PROBLEMS. 

I. 

(Sec  Chaplei  I  ) 

1.  How  many  ergs  are  equivalent  to  3.52  (18°)  calories? 

Ans.  147,241,600  ergs. 

2.  What  velocity  will  a  force  of  2.12  dynes  acting  for 
2  seconds  upon  a  body  weighing  3.62  grams  give  the 
body?  Ans.  1.17  cm.  per  second. 

3.  The  velocity  of  a  body  weighing  1.23  grams  is  4.3 
cm.  per  second,  what  force  has  acted   upon  it  for  one 
second?  Ans.  5.29  dynes. 

4.  What   is   the    pressure   in   dynes   per   sq.    cm.    of 
721  nim.  of  Hg?  Ans.  961,000. 

II. 

(See  Chapter  II.) 

5.  An  open  vessel  is  heated  to  819°  C.    What  por- 
tion of  the  air  which  the  vessel  contained  at  o°  remains 
in  it?  Ans.  0.25. 

6.  An  open  vessel  is  heated  until  one-half  of  the  gas 

453 


454  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

contained  at  15°  is  driven  out.     What  is  the  temperature 
of  the  vessel  ?  Ans.  303°  C. 

7.  A  volume  of  gas,  measured  at  15°,  is   50  cc.     At 
what  temperature  would  its  volume  become  44  cc.  ? 

Ans.  —  ig°.6  C. 

8.  A  volume  of  gas  at  766  mm.  pressure  is    137  cc. 
What  would  it  be  at   757  mm.?         Ans.  138.7  cc. 

9.  What  volume  does   i  mole  of  gas  occupy  at   50°, 
the  pressure  being  760  mm.  ?    At  100°,  p  being  900  mm.  ? 

Ans.  50°  =  26.5,   at   100°  =  25.8   liters. 

10.  A  volume  of  air  in  a  bell  jar  over  water  measures 
975  cc.     The  water  in  the  jar  is  68  mm.  above  the  water 
in  the  trough,   and  the  barometer  stands  at   756  mm. 
What  would  the  volume  be  if  exposed  to  standard  pres- 
sure, the  specific  gravity  of  Hg  being  13.6  ?    Ans.  963.4. 

11.  At    14°  C.    and    742    mm.    pressure   a   volume   of 
gas   measures    18  cc.     What   will   be   its   volume  at   o° 
and  760  mm.  pressure  ?  Ans.  16.72. 

12.  A  volume  of  H  at  a  temperature  of  15°  measures 
2.7  liters,  with  the  barometer  at  752  mm.     What  would 
have   been   its    volume   had    the   temperature   been   9°, 
and    the    pressure    762    mm.  ?  Ans.  2.6  liters. 

13.  What  volume  is  occupied  by  44  grams  of  oxygen  at 
70  cm.  Hg  pressure  and  35°  C.  ?         Ans.  37.7  liters. 

14.  J  mole  of  H,  \  mole  of  O,  and  \  mole  of  N  are 
mixed  in  a  volume  of  10  liters  at  o°  C.  and  760  mm. 
What  are  the  partial  pressures  of  H,  O,  and  N? 


PROBLEMS.  455 

Am.  11  =  1156.96,  0  =  1156.96,  and  N  =  77i.65  grams 
per  sq.  cm. 

15.  What  would   these  pressures    (14)    be  in  atmos- 
pheres at  10°  C.  ? 

Ans.  H  =  1. 164,  O  =  1.164,  and  N  =0.774. 

1 6.  i  liter  of  N  weighs  1.2579  grams  at  o°  and  760 
mm.     Calculate  the  specific  gas  constant,  r. 

Ans.  3307  gr.  cm. 

17.  The  specific   gas   constant,   r,   for  N  was  found 
above  (16).    What  it  is  for  H?    Atomic  weight  of  N 
is  14.073,  and  of  H  is  1.032.  Ans.  41,010  gr.  cm. 

18.  How  much  will  100  liters  of   chlorine   at  74  cm. 
Hg  pressure  and  30°  C.  weigh?          Ans.  278.7  grams. 

19.  A  solid  gives   off  a  gas  which  is  dissociated  to 
41%,  into  two  products.    What  is  the  work  done,  in 
calories,  gram-centimeters,  and  liter-atmospheres,  when 
i  mole  of  solid  goes  into  the  gaseous  state,  the  tempera- 
ture of  dissociation  being  55°  C.  ? 

Ans.  925  cals.,39,4io,coo  gr  cm  ,  37.96  L.  A. 

20.  How  much  work  will  be  done  by   i  kg.  of   CC>2 
when  heated  200°?  Ans.  373.1  L.  A.,  9088  cals. 

21.  The  time  necessary  for  CC>2  to  flow  through  a 
small  opening  is   26.5  minutes.     Under  the  same  con- 
ditions the  time  for  H  is  5.6  minutes.     Find  the  density 
of  CO2.  Ans.  22.3. 

22.  H  is  at  the  partial  pressure  of  2.136  atmospheres 
in  a  space  of  10  liters.     How  many  moles  per  liter  are 


45<>  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

there,    the    temperature    being    o°    and    the    pressure 
76cm.?  Ans.  c=  0.0954. 

23.  Starting  with  i  mole  of  A  in  22.4  liters  (at  o°,  760 
mm.  of  Hg),  assume  the  dissociation  according  to  the 
scheme  A  =  2B  +  $D  (here  A,  B,  and  D  represent  moles) 
to    be    23%.      What   will   be   the   final   volume,  where 
pressure  and  temperature  remain  unchanged? 

Ans.  43  liters. 

24.  What  is  the  final  concentration  of  A,  B,  and  D  in 
the  above  ? 

Ans.  A  =0.0179,  £=0.0107,  and  D  =  0.01604  moles  per 
liter. 

25.  The  molecular  weights  of  the  above  are  MA  =170, 
MB  =25)  and  MD  =40.     How  many  grams  per  liter  are 
there  of  each  at  equilibrium? 

Ans.  ^4=3.04,  B=  0.268,  and  D  =  0.642. 

26.  Assume  17  grams  of  A  (Af  =  170)  in  2.24  liters  (o°, 
760   mm.    Hg).     Find   concentrations,    partial   pressures 
and  grams  per  liter  of  A,  B,  and  D  where  the  dissociation 
of  A  is  20%  (M B  =2  5  >  M#=4.o),  and  the  volume  and 
temperature  remain  constant.     What  is  the  total  pres- 
sure of  the  system? 

Ans.  A  =0.036,  5  =  o.oi8,  17  =  0.027  moles  per  liter. 
^4=6.o6,    5=0.447,  Z)=i.7  gr.  per  liter. 
A  =0.8,      B  =0.4,      D  =0.6  atmospheres. 
Total  pressure  =  i  .80  atmospheres. 

27.  The  time  of  outflow  of  a   gas  is  21.4  minutes,  H 


PROBLEMS.  457 

being  as  in   (21)    5.6  minutes.     Find  molecular  weight 
of  the  gas.  Ans.  29.2 

28.  When  heated  PC^  dissociates  into  PCla  and  C\2- 
The  molecular  weight  of  PC15  is  208.28.     At  182°  the 
density  is  73.5,  and  at  230°  it  is  62.     Find  the  degree 
of  dissociation  at  182°  and  230°. 

Ans.  al82o=4i.7%,  a230o=68%. 

29.  Sulphur    molecules,    Sg,    dissociate    under  certain 
conditions  into  those  of  the  form  82.     If  this  dissocia- 
tion were  complete,  what  would  be  the  density  of  the 
gas  formed  of  the  82  molecules?  Ans.  32. 

30.  The  ratio  of  the  specific  heat  for  constant  pres- 
sure to  that  for  constant  volume  is  equal  to   1.67  for 
helium.     How   many  atomic   weights   are   there  in   the 
molecular  weight? 

31.  How  many  atomic  weights  are  there  in  the  molec- 
ular weight    of   argon,   the   specific    heat   for   constant 
volume  being  0.075  ?    The  density  gives   the  molecular 
weight  of  40.  Ans.  i. 

32.  What  is  the  specific  heat  of  CO2  gas  at  constant 
volume,  i.e.,  cv?    The  molecular  weight  is  44  and  the 
temperature  is  50°.  Ans.  0.164. 

33.  The  time  of  outflow  through  a  small  opening  is 
found  for  O  as   22.3  minutes;    for  H  it  is   5.6.     How 
many  atomic  weights  are  there  in  one  molecular  weight 
of  oxygen  ?  Ans.  2. 

34.  The  specific   heat   for   constant   pressure,   cp,   of 


45^  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

benzene  is  0.295,  what  is  the  specific  heat  for  constant 
volume,  cv?  Ans.  0.27. 

35.  The  specific  heat,  cv,  of  CO 2  is  0.2094;  what  is 
the  ratio  of  that  for  constant  pressure  to  that  at  con- 
stant volume  ?  Ans.  —  =#  =  1.22. 

V 

36.  CC>2  gas,  for  which  k  was  found  in  (35),  is  com- 
pressed in  a  flask  to  the  pressure  1.5  atmospheres,  and 
this  pressure  allowed  to  equalize  rapidly  against  the 
atmospheric  pressure,  equal  to  760  mm.  What  is  the 
temperature  produced?  The  original  temperature  was 
o°  C.  Ans.  -i9°.2C. 

37.  Air    (£  =  1.4)    compressed    to    pressure    equal    to 
50  atmosphere^,  is  expanded  rapidly  to  double  its  initial 
volume  against  the  pressure  of  the  atmosphere.     What 
temperature  is  produced  by  the  expansion?    The  origi- 
nal temperature  was  o°  C.  Ans.  -66°.i  C. 

What  effect  has  the  original  pressure  here? 

38.  Two    gases,    #1  =  1.4,    #2  =  1-22,    are     compressed 
rapidly,  each  to  TV  of  its  original  volume.     What  tem- 
peratures are  produced  if  o°  is  the  initial  one? 

Ans.    41=4i2°.8,  /fe2=i8o°.i  C. 

39.  Find  the  mechanical  equivalent  of  heat  by  Mayer's 
method,  using  O.     cv  for  O  is  0.2751;    i  gram  of  O  has 
the  volume  of  699.25  cc.  at  o°  and  760°  mm.  pressure. 


PROBLEMS.  459 


III. 

(See  Chapter  III.) 

40.  A  volume  of  50  liters  of  air  in  passing  through 
a  liquid  at   22°  causes  the  evaporation  of  5  grams  of 
substance,  the  molecular  weight  of  which  is  100.     What 
is  the  vapor-pressure  of  the  liquid  in  grams  per  square 
centimeter?  Ans.  25. 

41.  The  vapor-pressure  of  a  substance  is  44  mm.  at 
20°  C.,  the  molecular  weight  is  46.     How  many  grams 
of  liquid  will  evaporate  when  100  liters  of  air  pass  through  ? 

Ans.  11.08. 

42.  w-hexane  (^=85.82)  has  a  molecular  volume  in 
the  gaseous  state  equal  to  34,500  c.c.,  that  in  the  liquid 
state   being    137.96;    both   being   measured   at   60°   C. 
What  is  the  latent  heat  of  evaporation  of  i  gram  of  n- 
hexane?  Ans.  85.7  cals. 

43.  For   CC>2   at   o°   we   have   the   following   values: 
density  in  liquid  state  0.905,   density  in  gaseous  state 
0.099.     What  is  the  latent  heat  of  evaporation  of  i  gram 
of  liquid  CC>2?  Ans.  54.9.     Observed  is  57.48. 

44.  The  latent  heat  of  evaporation  of  chloroform  at 
61°  is  58.49  cals.,  the  boiling-point  is  61°,  and  the  molec- 
ular weight  in  gaseous  form  is  119.1.    What  is  the  for- 
mula of  the  substance  in  the  liquid  state  ? 

45.  The  latent  heat  of  methyl  alcohol  at  its  boiling- 


460  ELEMENTS  OF  PHYSICAL    CHEMISTRY. 

point,  66°,  is  267.48  cals.,  the  molecular  weight  in  gaseous 
form  being  32.     Find  molecular  weight  in  the  liquid  form. 
What  conclusion  is  to  be  drawn  from  this  result  ? 

46.  The  heat  of  evaporation  of  ether  at  34°. 5  is  88.39 
cals.    This  is  for  one  gram.     Find  the  change  in  boiling- 
point  due  to  a  change  from  760  to  634.8  mm. 

Ans.   -4°.76. 

dp 

47.  Find  heat  of  evaporation  of  ethyl  formate.    -7^  =  27 

mm.,  the  boiling-point  is  54°- 3  at  760  mm.,  and  the  molecu- 
lar weight  is  74.  Ans.  102.8. 

48.  Find     ,—    for    ethyl    propionate.     The    heat    of 

evaporation  is  77  cals.,  the  density  in  liquid  form  is 
0.7958,  and  in  the  gaseous  state  0.0033,  both  at  the 
boiling-point,  99°.  Ans.  21.5  mm. 

49.  The  refractive  index  of  liquid  N  is   1.2062,  and 
for  liquid  air  is   1.2053.     Find  that  for  liquid  O.     Air 
is  composed  by  weight  of  23.01%  of  O  and  76.99%  of 
N,  do  =  1.124,  ^=0.9673,  dw  =0.885.  Ans.  1.16. 

For  further   examples   on   refractive  index  see  May- 
bery  and  Shepherd.     (Am.  Chem.  Jour.  29,  278,  1904.) 

50.  What  is  the  surface  tension  in  dynes,  of    C6H6? 
The  radius  of  the  capillary  tube  is  0.01843  cm.,  at  46° 
the  height  of  the  liquid  is  3.213  cm.,  and  the  density  is 
0.85.  Ans.  #  =  24. 7  dynes. 

51.  Find  height  to  which  CS2  will  rise  in  a  capillary 


PROBLEMS.  461 

tube  at  46°,  r  being  equal  to  0.01708  cm.,  x  (the  sur- 
face tension)  being  27.68  dynes,  and  5  (the  density) 
=  1.224.  Ans.  2.7  cm. 

52.  Find  the  molecular  weight  of  C6H6  in  the  liquid 
state,   #  =  24.71    at   46°,    critical   temperature   is    288°. 5, 
M  in  gaseous  form=  78,  k=  2.12,  and  density=o.85. 

53.  Find    molecular   weight    in    liquid    form  of  CS2. 
#(46°)  =  27.68,  M  in  gaseous  form  =  76,  critical  tempera- 
ture is  28o°.3,  and  density  =1.224. 

54.  At    i4°.8>cetyl  chloride    (density  =  1.079)  ascends 
to  a  height  of  3.28  cm.  in    a    capillary   tube  of  which 
r  =  0.0142 5.     At  46°.2,  in   the  same  tube,  A  =  2.85    and 
the  density  is  1.064.     Find    the    critical   temperature  of 
acetyl  chloride,  ^=78.5.  Ans.  2^0.2  C. 

55.  What    is  the  heat  of  evaporation  of  ether  at  80° 
if  the  external  pressure  is  increased  so  as  to  increase 
the    boiling-point   from    35°    to    80°  ?     Specific    heat   of 
liquid  at  80°  is  0.690,  molecular  weight  is  74,  and  heat 
of  evaporation  at  35°  is  88.39   cals.  (see  data  in  text). 

Ans.  77.14  cals. 

HT 

56.  Find  -j-  for  ether  at  80°  as  given  in  (55);    the 

vapor-pressure  of  ether  at  80°  is  299.14  cm. 

Ans.  0.0146  per  mm. 

57.  What    external    pressure   is    necessary    to    change 
the  boiling-point  of  ether  in  (55)  from  35  to  80°  ? 

Ans.  4.06  atmos. 


462  ELEMENTS   OF  PHYSICAL   CHEMISTRY. 

58.  Find  molecular  weight  of  methyl  formate.     Critical 
temperature  is   214°  C.,   critical  pressure  59.25    atmos- 
pheres, and  critical  density  0.3494.         Ans.  M=6$.2. 

59.  Find   molecular  weight   of   ethyl  formate.     Criti- 
cal   temperature    is    235°.2    C.,    critical    pressure    46.83 
atmospheres,  and  critical  destiny  0.3232.  Ans.  If  =  77.2. 

60.  What  is  the  change  in  vapor  pressure  for   ether 
(56)  per  degree  ?  Ans.  68.5  mm. 

IV. 

(See  Chapter  IV.) 

.  61.  The  specific  heat  of  Ni  is  0.1092.  What  is  the 
atomic  weight  of  Ni?  Ans.  58. 

.  62.  The  specific  heat  of  Fe  is  0.112.  Find  atomic 
weight  of  Fe.  Ans.  56.6. 

63.  The  specific  gravity  of  solid  phenol  is  1.072,  and 
in  the  liquid  state  it  is  1.0561;   the  latent  heat  of  fusion 

jrr< 

is  24.93,  and  the  melting-point  is  34°.     Find  j-. 

Ans.  0.00421°  per  atmos. 

dT 

64.  Acetic  acid  melts  at  i6°.6  C.,  j~  =o°.o242  per  at- 
mos., the  heat  of  fusion  is  46.42  cals.     Find  change  in 
volume  by  the  liquefaction  of  i  gm.     Ans.  0.00016  liter. 

,  65.  The  specific  heat  of  silver  sulphide  is  0.0746,  why 
is  the  formula  regarded  as  Ag2S? 

66.  Specific  heat  of  solid  acetic  acid  is  0.4599  fr°m 
io°  — 15°,  of  fluid  is  0.5026.  Find  heat  of  fusion  at  50° 
from  that  at  5°.6,  which  is  44.34  cals.  Ans.  46.24. 


PROBLEMS.  463 

67.  Naphthalene    in    solid    state    has    specific    heat 
=0.356,  in  liquid  state    =0.396,  both  at  80°,  and  the 
heat  of  fusion  is  ^'=35.5  cals.     What  is  w  at  30°? 

Ans.  33.5. 

68.  Assuming  the  data  given  in  67,  find  the  fraction  of 
naphthalene  which  would  separate  as  the  result  of  an 
overcooling  of  10°.  Ans.  0.112. 

69.  The  vapor-pressure  of  both  ice  and  water  at  o°  is 
4.62  mm.,  the  heat  of  fusion  of  ice  is  80  cals.     What  is 
the   difference  between   the   temperature   coefficients   of 
vapor-pressure  for  the  two  states? 

Ap     AP 

Ans.  0.0446.     Observed    is  -7^ --77^  =0.3852 -0.3399 

=  0.0453  mm. 

70.  The  heat  of  fusion  of  ice  is  80  cals.  and  the  heat  of 
evaporation  of  water  is  597  all  at  o°.     What  is  the  heat 
of  sublimation  of  ice  at  o°?  Ans.  677  cals. 

71.  How  could  the  heat  of  sublimation  of  ice  at  o° 
(see    above)    be    determined    in    another    way?    What 
value  would  it  give?     The  specific  volume  of  saturated 
water-vapor  at  o°  is    204,680  (i.e.  p  =4.619    mm.  Hg), 

that  of  ice    is    1.99   cc.   and   -j^  for  ice  is  0.3852  mm. 

Ans.  687  cals. 

72.  The  specific  gravity  of  water  at  o°  is  i,  the  heat 
of  solidification   is   80.     What   change  in   the  freezing- 
point  of  water  results  from  an  increase  of  pressure  of 
i  atmosphere?    (For  data  see  above.)    Ans.  —0.00748. 


464  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

73.  The  specific  in   the  liquid  state  is  0.43,   in  the 
solid  state  0.352.     An  overcooling  of  i7°.8  causes  0.25 
of  the  liquid  to  separate  as  solid.     What  is  the  latent 
heat  of  fusion  at  the  freezing-point?    What  is  it  at  10° 
below  the  freezing-point?    Ans.  30.6  and  29.84  cals. 

74.  The  specific    heat    of   phosphorus   in    the    liquid 
state  at  44°  is  0.204,  the  latent  heat  is  5  cals.,  and  at 
39°. 5  is  4.86  cals.     What  is  the  specific  heat  in  the  solid 
state?  Ans.  0.173. 

75.  At  29°  the  latent  heat  of  fusion  of  CaCl2-6H2O  is 
40.7,  and  at   — 160  is  o.     What  is  the  specific  heat  in 
the  liquid  state,  that  in  the  solid  state  being  0.345? 

Ans.  0.560. 
V. 

(See  Chapter  V.) 

.  76.  One  liter  of  H2O  absorbs  i  liter  of  CO2  at  o°,  760  mm. 
How  many  grams  of  CO2  gas  are  contained  in  a  bottle  of 
carbonic  water  holding  350  cc.  of  solution,  the  pressure 
being  5  atmospheres?  Ans.  3.44  grams. 

77.  Air  is  composed  of  20.9  volumes  of  O  and  79.1 
of  N.  At  15°  water  absorbs  0.0299  volumes  of  O  and 
0.0148  of  N,  the  pressure  of  each  being  that  of  the  at- 
mosphere. What  is  the  composition  of  air  absorbed  in 
H2O  ?  Ans.  34.8%  by  volume  of  O  and  65.2  of  N. 
•  78.  Find  molecular  weight  of  naphthalene;  49.4  grams 
of  H2O  and  8.9  grams  of  naphthalene  go  over  when  dis- 
tilled at  98°.  2  and  733  mm.  The  vapor-pressure  of 
H2O  at  98°.2  is  712.4.  Ans.  112. 


PROBLEMS.  465 

79.  Supposing   that   the   air   dissolved   in    (77)    were, 
collected  and  redissolved  in  water,  what  would  be  the 
composition  of  the  air  dissolved? 

Ans.  51.9%  by  volume  of  o  and  48.1  of  N. 

80.  What  is  the  surface  energy  involved  on  a  cubic 
centimeter  of  gypsum  powder  (2.5  grams)  when  it  is  re- 
duced to  cubical  particles  with  an  edge  equal  to  o.oooi 
mm.    and    added    to    water?    The    surface    tension    of 
water  is  80  dynes.  Ans.  4.8  X  io7  ergs. 

81.  What  is  the  osmotic  pressure  of  a  i%  solution  of 
glucose  (M  =  1 80)  at  o°  C.  ? 

Ans.  94.6  cm.  H.     Obs.  =94  cm. 

82.  The  osmotic  pressure  of  a  solution  of  cane-sugar 
at  o°  is   49.3   cm.   of  Hg.     What  percentage  of  sugar 
(M  =342)  is  contained  in  it  ?   Ans.  0.99% ;  obs.  =  1.0%. 

83.  The   molecular  weight  of   H^O   is    18,   of  nitro- 
benzene   is    121.     Vapor- pressure    at    99°  '  (the   boiling- 
point  of  the  mixture)  of  pure  H^O  is  733  mm.,  and  of 
nitrobenzene  is  27  mm.     How  much  nitrobenzene  is  con- 
tained in  the  distillate?         Ans.  0.2  of  total  by  weight. 

*  84.  The  osmotic  pressure  of  a  sugar  solution  at  32°  C.  . 
is  54.4  mm.  What  is  it  at  i4°.2?  Ans.  51.2  mm. 

x  85.  The  osmotic  pressure  of  solution  containing  io  * 
grams  of  sugar  to  a  certain  volume  is  200  mm.  What  \ 
is  that  for  the  same  volume  containing  13.5  grams? 

Ans.  270  mm. 
86.  10.442  grams  aniline  in  100  grams  of  ether  gives  a 


466  ELEMENTS  OP  PHYSICAL  CHEMISTRY. 

vapor-pressure  of  210.8  mm.     Ether  alone  (Af  =  74)  gives 
229.6.     Find   molecular   weight  of  aniline.     Ans.  87. 

87.  Find  osmotic  pressure  at  o°  of  aniline  in  (86)  in  at- 
mospheres and  gram  per  square  centimeter,     s  for  ether 
is  0.737.        Ans.   19.82  atmos.  or  20460  gr.  per  sq.  cm. 

88.  The  osmotic  pressure  of  a  substance  in  water  solu- 
tion is   100  cms.  at  o°  C.    Find  the  vapor-pressure  of 
the  solution;  that  of  water  at  o°  is  4.57  mm. 

Ans.  4.56  mm. 

89.  What  is  the  work,  in  gr.-cms.  and  calories,  neces- 
sary to  separate  200  grams  of  a  substance  (M  =66)  from 
the  solvent  at  20°  C.  ?     10  grams  of  substance  to  the  liter 
of  solvent.  Ans.  1953  cals.;  83,200,000  gr.  cms. 

90.  64.74  grams  of  propylene  bromide  (If  =  222)  are 
mixed  with  145.93  grams  of  ethylene  bromide  (M  =  206) 
at  85°.o5;  the  vapor-pressure  of  pure  propylene  bromide 
at  this  temperature  is  127.2  mm.,  and  that  of  ethylene 
bromide  is  172.6  mm.     What  is  the  partial  vapor-pres- 
sure of  each  in  the  mixture?    What  is  the  composition 
of  the  distillate?    What  is  the  total    vapor-pressure  of 
the  mixture  ? 

Ans.  f     (ethylene  bromide)  =122.3  mm> 

observed  =  121. 4    " 

p'  (propylene  bromide)  =   37.1     " 

observed  =  37.3     " 

total  vapor-pressure  =  1 59.4    ' ' 

observed  =  158. 7    " 


PROBLEMS.  467 

24.6  grams  of  propylene  bromide  and  75.4  grams  of 
ethylene  bromide  in  each  100  grams  of  distillate.  Ob- 
served values  are  24.9  and  75.1. 

91.  The  increase  in  the  boiling-point  of  54.65  grams  of 
CS2  caused  by  the  addition  of  1.4475  gr&ms  of  P  is  o°.486. 
What  is  the  molecular  weight  of  P  in  CS2  ? 

9Ans.   129.2. 
What  is  the  formula,  the  atomic  weight  being  31  ? 

92.  Calculate   the   increase   in   boiling-point   of   ether 
when  to  100  grams  we  add  a  mole  of  a  substance.    The 
boiling-point  of  ether  is  34°.97,  the  latent  heat  of  evapo- 
ration is  88.39.  Ans.  2i.°5. 

93.  The   molecular   increase   of   the   boiling-point   of 
H2O,  as  caused  by  the  addition  of  i  mole  of  substance 
to  100  grams,  is  5°.2.     Find  heat  of  evaporation  of  H2O. 

Ans.  535.1  cals. 

Ap 
4    94.  At  40°  ~j=*  for  benzene  (M  =  78)   is  0.88 1  for  a 

mean  pressure  of  22.42  cm.  What  is  the  molecular 
increase  of  the  boiling-point  of  benzene  under  this  reduced 
pressure?  Ans.  i9°.85. 

95.  10  grams  of  a  substance  in  100  grams  of  a  solvent 
increase  the  boiling-point  by  o°.87.    The  molecular  weight 
of  the  substance  is  60.     Find  the  molecular  increase  of 
the  boiling-point.  Ans.  5.22. 

96.  0.284  grams  of  the  oxime  (CH3)2CNOH  causes  a 
decrease  of  o°.i55  in  the  freezjng-point  of  100  grams  of 


468  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

glacial  acetic  acid.  K  for  acetic  acid  is  38.8.  Find 
molecular  weight  of  the  oxime.  Ans.  71. 

.  97.  The  ionization  of  a  normal  solution  is  80%,  two 
ions  being  formed.  What  will  the  depression  of  the 
freezing-point  be>  water  (£  =  18.9)  being  the  solvent? 

Ans.  3°.4. 

98.  The  molecular  depression  of  an  aqueous  solution 
containing  an  ionized  substance  is  22°.     Find  the  degree 
of  ionization  of  the  substance  in  that  volume. 

Ans.  a  =16.3%. 

99.  A   normal  water  solution   gives   a   depression   of 
2°. i,  when  the  overcooling  is  2.21.     What  is  the  strength 
of  the  solution  whose  freezing-point  is  observed? 

Ans.  1.028  N. 

100.  Find  the  freezing-point  depression  of  the  normal 
solution  itself  in  (99).  Ans.  2°.O4. 

.  IQI.  Find  the  heat  of  fusion  of  nitrobenzene;  its  melt- 
ing-point is  5°.3,  the  molecular  depression  of  the  freezing- 
point  is  #  =  70.7.  Ans.  21.9  cals. 

102.  The  specific  heat  of  nitrobenzene  is  0.3524,  its  heat 
of  fusion  is  21.9  cals.  Find  the  amount  of  solid  separated 
by  an  overcooling  of  i°.         Ans.  16.1  grams  per  1000. 

103.  In  (91)  find  the  osmotic  pressure  at  46°  of  P  in 
the  CS 2  solution.     sCSa  =  1.2224.  Ans.  6.84  atmos. 

104.  Find  the  vapor-pressure  in  (91)  of  P  in  CS2  solu- 
tion at  o°  the  vapor-pressure  of  CS2  at  o°  is  127.91  mm. 

Ans.  125.9  mm- 


PROBLEMS.  469 

105.  In  (96)  find  the  osmotic  pressure  of  the  oxime  in 
glacial  acetic  acid  at  17°,  sp.  gr.   (acetic  acid)  =1.056? 

Ans.  i  atmos. 

106.  What  is  the  relation  between  the  osmotic  pres- 
sures of  .01  mole  of  substance  in  a  1000  grams  of  water 
and  a  1000  grams  of  ether  (sp.  gr.  =0.7370),  assuming  the 
same  molecular  state  of  the  solute  in  each  ? 

Ans.  P,=o.737oPw. 

107.  In   (96)   find  the  vapor-pressure  of  the  solution 
at  40°  C.,  the  vapor-pressure  of  glacial  acetic  acid  at 
40°    being    34.77    mm.  Ans.  34.69  mm. 

108.  A   o.i   normal  solution  of  acetic  acid  in  water 
freezes  o°.i927  lower  than  H^O.     Find  the  degree  of 
ionization    of    the    acetic    acid.  Ans.  a  =  2%. 

109.  A   .15   normal  solution   of  succinic  acid  freezes 
o°.2864  lower  than  H2O.     Find   the'  ionization   of  the 
acid.  Ans.  a  =  i%. 

no.  A  saturated  aqueous  solution  of  ether  freezes  at 
—  3.85.  When  4.76  grams  of  I  are  dissolved  in  100  grams 
ether,  a  saturated  aqueous  solution  freezes  at  —3.789. 
Find  molecular  weight  of  I  in  ether.  The  molecular 
increase  of  the  freezing-point  of  water  by  the  addition 
of  i  mole  of  substance  to  the  ether  is  3°.o6  over  that  point 
for  pure  ether  in  water  (see  pp.  190,  191).  Ans.  239. 


47°  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

VI. 

(See  Chapter  VI.) 

in.  Find  the  heat  of  decomposition  of  H2O2  in  water 
from  the  following  reactions: 


SnCl2  •  H2Cl2Aq  +  H2O2Aq  =  SnCl4Aq  +  2H2 

Ans. 

112.  The  heat  of  combustion  of  NH3  in  O  at  constant 
volume  is 

2NH3  +  30  =  3H2O  +  N2  +  181  2K. 

Find  heat  of  formation  of  NHs  from  its  elements. 

Ans.  io6.2K. 

113.  How  does  the  heat  of  combustion  of  H  with  O 
to  form  1  8  grams  of  liquid  H2O  at  constant  volume  vary 
with  the  temperature?     The  heat  capacity  of  72  grams 
of  H2O  is  0.72  K,  and  that    of  4  grams  of    H   plus  32 
grams  of  O  is  0.2056.  Ans.   —0.0772^  per  degree. 

114.  C#4  +  2O2=CO2  +  2H2O  +  2ii9#     at     constant 
pressure  and   18°  C.     Find  heat  of  formation  of  CH4. 

Ans.  2iS.2K. 
115. 


PROBLEMS.  47  * 

Find  the   heat    formation  of    KOHAq    from  the  ele- 
ments. Ans.  n6$K. 

116.  Zn  +  2HCLAq=ZnCl2Aq  +  2H  +  342/£  at  constant 
pressure.     What  is  it  for  constant  volume?    Tempera- 
ture is  20°.  Ans.  347.9^. 

117.  The  following  reactions  take  place: 


C6H6  +  150  =6C02  +3H20  +yK 
and 

C6H6  +  750  =  6CO2 


At  27°,  assuming  these  to  be  at  constant  pressure,  find 
values  for  constant  volume. 

1  1  8.  What  is  the  difference  in  energy  between  18  grams 
of  water  at  a  100°  and  water-  vapor  at  the  same  tem- 
perature? Between  water  and  ice  at  o°?  The  latent 
heat  of  evaporation  is  536.4,  of  fusion  is  80  cals. 

Ans.  9655  and  1440  cals. 

119.  Find  heat  of  formation  of  gaseous  hydrobromic 
acid  from  the  reactions, 

SO2Aq  +  O  =SOsAq-f  63700  cals., 
2Br  +  SO2Aq  +  H2O  =  2HBrAq  +  SO3Aq  +  54000  cals., 
when,  in  addition,  we  know  that 

H2  +  O  =H2O  +  68400  cals., 
and  H  Br  ^HBrAq  +  20000  cals. 

Ans.  H+Br=HBr+93$o  cals, 


47  2  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

1  20.  Find  the  heat  of  formation  and  heat  of  forma- 
tion in  solution  of  NHa. 

+  30  =N2  +  3H2O  +  181200  cals.  j  const. 
+  30  =  3H2O  -f  3  X  68400  ]  pressure. 


Ans.  N  +  $H  =  12,000  cals.  ;  (N,  3H",  aq)  =  20,400  cals. 

121.  NaOH  +  5oH2O   has   a   molecular   specific   heat 
equal  to  885.0  cals.  and  SO3  +  iooH2O  that  of  1797  cals. 
At  9°.i6  these  evolve  32,060  cals.  by  combining,  and  at 
24^42  they  generate  31,650  cals.     What  is  the  specific 
heat  (i.e.,  for    i     gram)    of    a    solution   of    Na2SO4+ 

2ooH2O  ? 

Ans.  0.9603  cal. 

122.  What  is  the  heat  of  formation  of  a  very  dilute 
solution  of  magnesium  chloride?     (See  text  for  data.) 

Ans.  1875  cals. 

VII. 

(See  Chapter  VIII.) 

123.  In  the  volume  of   i  liter  there  are  0.14  mole  of 
hydrogen  and  0.081  mole    of  iodine.     At  the  tempera- 
ture of  440°  C. 


#1 

Find  the  amount  of  hydriodic  acid  formed. 

Ans.  0.14855  mole. 


PROBLEMS.  473 

124.  The  initial  pressure  of  I  is  38.2  cm.,  the  fraction 
uniting  with  H  is  0.8.     What  is  the  original   pressure 
of  the  H?    K=o.o2.  Ans.  40.35    cm. 

125.  The    coefficient    of    distribution  of    acetic     acid 

0.245        ,   0.314 

between  water  and  benzene  is  as   -       -   and at 

0.043  0.071 

two  dilutions.    What  is  the  molecular  weight  in  benzene  ? 
In  water  it  is  60.  Ans.  2.02X60. 

126.  Find  the  constant  of  equilibrium,  for  concentra- 
tions, for  the  reaction  2NO2  =  2NO  -fO2  at  279°  from 
the  following  data,  du  =  1.590  X  2 : 


t 

yn 

4 

a 

130° 

718-5 

1  .600 



184° 

754-6 

J-551 

0.050 

279° 

737-2 

1-493 

0.130 

494° 

742-5 

1.240 

0-.565 

620° 

760.0 

i.  060 

I.OOO 

Ans.  ^279°  =  0.136  for  2  moles  NO2,  using  the  volume 
per  cent  of  each  in  the  formula,  i.e.,  parts  of  NO2t  NO, 
and  O  per  100  volumes,  viz.  81.67,  12.21,  and  6.1  respec- 
tively. 

127.  At  440°  in  50  liters  we  have  a  mixture  of  2.74 
moles  of  HI,  0.5  mole  of  H  and  0.3011  mole  of  I.     K 
=0.02.  In  which  direction  does  the  reaction  go? 

128.  At  440°  (#=0.02)  5.30  cc.  of  H  are  mixed  with 
7.94  cc.  of  iodine- vapor.     How  much  HI  is  formed? 

Ans.  9.475  cc.    Observed,  9.52  cc. 

129.  At  3000°  a  for  CO2,  according  to   the  formula 

2,  is  equal  to  0.4.    What  is  the  constant 


474  ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

of  dissociation  of  i  mole  of  CO 2  calculated  from  the 
volume  of  each  in  100  volumes?  Ans.  2.72. 

130.  What  is  K  (129)  for  2  moles  of  CO 2? 

Ans.  7.4  for  the  concentrations  expressed  in  parts  per 
hundred,  i.e.,  50  of  CO 2,  33  of  CO  and  17  of  O. 

131.  The  constant  of  equilibrium  for  the  reaction  of 
amylene  and  acetic  acid  is  830.1  (p.  249).    At  the  tem- 
perature at  which  this  is  true   2    moles  of  amylene  are 
mixed  with  a  moles  of  acid  in  2  liters,  and  0.5  mole  of 
ester  is  formed.     Find  a,  the  original  number  of  moles  of 
acid.  Ans.  0.50241  mole. 

132.  6.63  moles  of  amylene  with  i  mole  of  acid  shows 
that    0.838   mole    are    formed   in   the    total  volume   of 
894  liters.     How  much  will  be  found  when  we  start  with 
4.48  moles  of  amylene  and  i  of  acid  in  the  volume  of 
683  liters?  Ans.  0.8 1 1 1  moles. 

133.  Write  the  application  of  the  law  of  mass  action 
for  the  reaction 

Fe  +  4H2O  =  Fe2O4  +  4#2, 

and  calculate  K  for  H2O=4.6  mm.,  £[  =  25.8,  and  for 
H2O=io.i  and  11  =  5.79. 

134.  For  NH4HS=#2S  +  AT£r3,  #=62,400  (for  pres- 
sures in  mm.)  at  25°.!  C.     In  a  vacuum  at  25°.!  we  intro- 
duce NH$  and  H%S  until  we  have  a  partial  pressure 
of  the  former  of  300  mm.,  and  of  the  latter  594  mm. 
Then  the  reaction  is  allowed  to  take  place.     How  much 
does  each  gas  lose  in  pressure?  Ans.  157.2  mm. 


PROBLEMS.  475 

135.  At  i8°.4  i  mole  of  BaSC>4  dissolves  in  50,055  liters 
at  37°. 7  in  31,282  liters.     On  the  justified  supposition 
that  BaSC>4  is  completely  ionized,  find  the  heat  of  disso- 
ciation. Ans.  —8511  cals. 

136.  The  salt  Na2HPO4.i2H2O  has  a  vapor-pressure 
at  15°  of  8.84  mm.,  at  i7°.3  ^  =  10.53  mm.     Find  the 
heat  of  vaporization,  i.e.,  the  heat  change  during  loss 
of  i  mole  of  water  of  crystallization  by  evaporation. 

Ans.  —12,728  cals. 

137.  A  mixture  of  alcohol  and  hydrochloric  acid  is 
an  equilibrium  in  which  a  certain  amount  of  H2O  is 
formed.     At  77°,  #=0.307,  at  99°,  #=0.177.     Find  the 
heat  generated  by  the  reaction.  Ans.  +6527  cals. 

138.  The  speed   constant   of   formation   of   HI   from 
H  and  I  is  0.00023;  K,  the  equilibrium  constant  at  that 
temperature,  374°,  is  0.0157.    What  is  the  speed  con- 
stant of  decomposition?        Ans.  0.0146.  Obs.  0.0140. 

139.  To  i  liter  of   a  molar  solution  of  a  monobasic 
acid  (#=0.000018),  a  salt  with  an  ion  in  common,  having 
an  ionization  at  that  dilution  equal  to  100%,  is  added. 
How  much  in  moles  in  the  dry  state  must  be  dissolved 
to  decrease  the  concentration  of  H"  ions  of  the  acid  to 
o.i  of  its  previous  value?  Ans.  0.04211  mole. 

140.  What  difference  would  be  observed  if  this  had 
been  a  dibasic  acid  with  the  same  constant? 

141.  A  small  amount  of  a  base  is  mixed  with  a  large 


476  ELEMENTS  Oh  PHYSICAL   CHEMISTRY. 

amount  of  a  solution  containing  equi-molecular  amounts 
of  acetic  and  lactic  acids.  In  what  proportion  will  they 
form  salts?  Ans.  Acetic: lactic:  10.00424 -.0.0117. 

142.  A  small  amount  of  base  is  mixed  with  a  large 
amount  of  an  equi-molecular  mixture  of  two  acids.     The 
ionization  of  one  is  10%,  of  the  other  90%.    How  much  of 
each  salt  will  be  present  in  100  parts  of  the  salt  formed  ? 

143.  The  constant  for  the  first  hydrogen  of  malonic 
acid  is   i58Xio~5,  of  the  second  is  i.Xio"6.   What  is 
the*  amount  of  H'  ions  in  a  solution  of  i  mole  of  malonic 
acid  in  2000  liters?      Ans.  0.8385,  i.e.,  CHA/  =0.7985. 

144.  The  degree  of  ionization,  a  for  KC1  is  0.75  for 
V  =  i,  0.94  for  F  =  io,  0.99  for  F  =  10,000.     Calculate  K 
by  the  dilution  laws  of  Ostwald,  Rudolphi,  and  van't  Hoff. 

145.  The  value  of  K  for  a  0.05  molar  solution  of  acetic 
acid  is  0.0000175  at  18°,  and  0.000016*24  at  52°.     What 
is  the  heat  of  dissociation  of  acetic  acid  ?    At  what  tem- 
perature would  this  value  be  true?  Ans.  416  cals. 

146.  Find  the  heat  of  neutralization  of  i  mole  of  acetic 
acid  (in  200  moles  of  H^O)  with  i  mole  of  sodium  hydrate 
(in  200  moles  of  EbO)  at  35°;  a  for  acetic  acid  is  0.009, 
the  heat  of  ionization  is    —386  cals.;    a  for  NaOH  is 
0.861,  its  heat  is  — 1292  cals.,  and  a  for  sodium  acetate 
is  0.742,  its  heat  being  —391  cals.     The  heat  of  ioniza- 
tion of  water  at  35°  is  12,632  cals.      Ans.  13,093  cals. 

147.  The  solubility  of  barium  sulphate  at  3  7°.  7  is   i 


PROBLEMS.  477 

mole   in    37,282   liters.     What   is   its  solubility  product 
at  this  temperature?  Ans.  7.2Xio~10. 

148.  PbI2  is  soluble  to  0.00158  mole  per  liter  at  25°.2, 
i.e., 


What  is  its  solubility  in  presence  of  a  o.i  molar  solu- 
tion of  I'  ions  ?  Ans.  1.58  X  io~6  moles  per  liter. 

149.  MCN  (solubility  is  0.02,  and  is  ionized  completely) 
is   dissociated   hydrolytically   in   solution.     K  for  HCN 

=  i3Xio-10,  and  K  for  H2O  (25°)  =(i.o9Xio-7)2. 
Find  amount  of  M  ions  from  an  exterior  source  necessary 
to  be  in  solution  in  order  to  prevent  hydrolysis. 

Ans.  0.02162  mole  per  liter. 

150.  Water  under  atmospheric  pressure  at  16°  absorbs 
0.9753  liters  of  CO2  to  the  liter,  i.e.,  0.04354  moles  per 
liter.     Of  this  0.000115  mole  is  ionized   (0.264%)   mto 
H*   and   HCCV   ions.    What  is   the  solubility   product 
of  H2CO3?  Ans.  1.32  X  io-g. 

151.  At    25°  solubility  product  of  H2CC>3  into  H"  and 
HCO3'    is    1.32  Xio~8.     Find  concentration  of  H'  and 
HCO3'  ions. 

152.  The  solubility  product  of  the  substance  AC2  is 
0.00621.     Find   concentration  of   A   and    C   ions   when 
the  substance  ionized  completely  into  A"  and  2C'. 

Ans.  0.1157  mole  per  liter  of  A"  and  0.2314  of  C". 

153.  AgCNO   is    soluble   at    100°  to  0.008  mole  per 


ELEMENTS  OF  PHYSICAL   CHEMISTRY. 

liter.     How  much  would  dissolve  in  a  solution  contain- 
ing o.i  mole  of  Ag  ions? 

Ans.   =6.4X  io~4  moles  per  liter. 

154.  The  solubility  of  a  salt  is  o.oooi  mole  per  liter. 
The  formula  of  the  base  is  M(OH)4,  and  its  solubility 
is  o.ooooi,  ionizing  into  M::  and  4<DH'.  Will  it  dissociate 

hydrolytically  in  water  at  25°,  for  which  K  =  (1.09  X  io~7)2  ? 

Ans.  Yes. 

155.  At  a  dilution  of  32  liters  a  binary  substance  is 

0.9%  hydrolyzed.     What  is  the  percentage  of  hydrolysis 
at  loo  liters?  Ans.  1.584%. 

156.  The  solubility  of  Mg(OH2)  is  0.00002  mole  per 
liter.    If  there  are  1.5  moles  of  NH4  arid  o.oooi  of  a  mole 
of  Mg"  ions  present  in  a  molar  solution  of  NH4OH,  what 
concentrations  of   OH'    ions   must  be  added  before  the 
solubility  product  of  Mg(OH)2  can  be  reached?     K  for 
NH4OH=  0.000023.  Ans.  o.26Xio~5. 

157.  The  constant  of  hydrolytic  dissociation  at   100° 


( 
i.e.,K 


a\y''  What 
is  the  ionization  constant  of  NH4OH?  K(R2O  at  100°) 
=  (8.5Xio~7)2.  Ans.  214X10-7. 

158.  A  carbonate  MCO3  is   soluble    to    o.oooi    mole 
per  liter,   but  is   50%   hydrolytically    dissociated.     Find 
solubility  product.  Ans.  s  =  5.  X  io~9. 

159.  At  25°  the  solubility  of  AgBr  is  86Xio~8,  and  of 
Agl  is  0.97  Xio~8.     How  much  Br'   in   the   ionic  state 
must    be    added   to  start  precipitation   of   AgBr   from 
solution  of  Agl  ?  Ans.  7.63X10"  5. 


PROBLEMS.  479 

160.  AgCl,    AgBr,    and    Agl    are   dissolved    together. 
What  concentration  exists  of  Ag',  BrJ  Cl^  and  I'  ions  per 
liter?    The  solubility  of  AgCl  is  i.25Xio~5,  the  others 
are  given  above. 

161.  Brom-iso-cinnamic    acid    at    25°    is    soluble    to 
0.0176  mole  per  liter,  and  is  ionized  to  the   extent   of 
1.76%  into  H*  and  a  negative  ion.     What  is  the  solu- 
bility product  of  the  acid?  Ans.  g.6Xio~8. 

162.  How  much  of  this  acid  must  always  remain  in 
solution  at  25°,  even  in  the  presence  of  an  infinite  amount 
of  H*  ions  from  another  acid? 

Ans.  0.0173  mole  per  liter. 

163.  What  is  the  solubility  of  brom-iso-cinnamic  acid 
in  the  presence  of  a  o.ooi    molar  solution  of  H*  ions 
from  another  acid  ? 

Ans.  Solubility  =0.01 73 +0.0000883  mole  per  liter. 

164.  K  for  cinnamic  acid  at  25°  is  0.0000345,  and  its 
solubility  in  water  is  0.00331  mole  per  liter,  the  ioniza- 
tion  being  9.84%.     In  a  o.oi  molar  solution  of    aniline 
the   solubility   is   increased    to   0.00804    rnole   per  liter. 
The    salt    formed    is    ionized    at   this  dilution   to   93% 
and    sH2o    is    (1.09  Xio~7)2.       What    is    the    constant 

for  aniline?  Ans.  5.3  Xio"11. 


480  ELEMENTS  OF  PHYSICAL  CHEMISTRY. 

VIII. 

(See  Chapter  IX.) 

165.  Change  1000  calories  into  watt-seconds. 

Ans.  4178. 

166.  Change  350  watt-seconds  into  calories. 

Ans.  83.8. 

167.  An  aqueous   solution   of   CuSO4   is   electrolyzed 
until  0.2955  gram  of  Cu  are  deposited,  using  inert  elec- 
trodes.   The  liquid  at   the  cathode  before  passage  of 
the  current  gave   2.2762  grams  Cu,  and   after   passage 
2.0650  grams.     Find  the  mobilities  of  Cu"  and  SO4". 

Ans.  £7Cu..  =0.285,  Us04"=so-7iS- 

168.  An  aqueous  solution  of   CuSO4  is   electrolyzed 
with  Cu  electrodes  until  0.2294  gram  of  Cu  is  deposited. 
Before  electrolysis  the  solution  at  the  cathode  contained 
1.195  grams  Cu,  after  electrolysis    1.360    grams.     Find 
mobilities  of  Cu"  and  SO4". 

Ans.  UCu-  =0.28,  Z7So4"=o.72. 

169.  A  0.02  molar  solution    of    KC1   (icl8o  =0.002397) 
gives  in  a  certain  electrolytic  cell  a  resistance  of  150 
ohms.     Find   factor  which  will   transform   conductivity 
results  for  this  cell  into  specific  conductivities. 

Ans.  0.36. 

1 70.  Find  the  ionization  constant  for  NH4OH.   A8  =  3.4, 
^16=4.8,  ^32=6.7,  J64  =9.5,  Ji28  =  13.5  and  A25Q  =  18.2. 
A^   for  NH4C1    is    130.1,  A^  for    KOH  is  239.3,  and 
Aw  for  KC1  is  131.2.         Ans.  K  (average)  =0.000026. 

171.  At   18°  lactic  acid   gives   the   following    values: 


PROBLEMS.  481 

^32=23.11,    ^64  =  32.21,  Ji28=44-47>   4256  =  60.23,    4,72  = 

82.2,  and  ^1024  =  116.9.  What  is  the  ionization  con- 
stant? A^  for  HC1  is  383.9,  A^  for  NaCl  is  in,  and  An 
for  the  sodium  salt  of  lactic  acid  is  85.1. 

Ans.  K  (average)  =  0.000136. 

172.  In  a  o.oi  molar  solution  of  KNO3,  NO3'  has  a 
mobility  of  0.497,  and  K*  one  of  0.503.     Find  the  equiv- 
alent conductivity  of  NOa'   and    K*    in    this    solution, 
K=o.ooi044.  Ans.  /K-  =  52.5,  /NO3-=  51.9. 

173.  At   infinite  dilution  the    equivalent    conductivity 
of  a  solution  is  91.    What  would  it  be  at  a  dilution  at 
which  it  is  50%  dissociated  into  2  ions?    Ans.  45.5. 

174.  For  hydriodic  acid  we  have  the  following  data: 

^2=364*  ^4=376,  4*=384>  4e=39I>  4*2=397*  4*4 
=  402,  ^123=405,  and  ^256  =406.  Compare  the  con- 
stancy of  the  Ostwald  and  van't  Hoff  dilution  laws. 


175.  The  velocity  of  migration  of  Ag*  is  0.00057  cm- 
per  second,  as  is  also  that  of  ClOs'.     What  is  the  equiva- 
lent conductivity  of  a  solution  of  AgClOa  which  is   infi- 
nitely dilute  ?  Ans.  no. 

176.  The  equivalent  conductivity  of  the  sodium  salt 
of  an   acid   is    ^32=89.9,   ^54=97.1,  ^123  =  104.5,   ^255 
=  in.  i,   ^512  =  117.2,   and  ^1024  =  122.7.     What   is   the 
basicity  of  the  acid  ? 

177.  The    equivalent    conductivity    of   a    solution    of 
Na2SC>4  in  256  liters  at   25°  is   141.9.    What  is  it  at 
infinite  dilution?  Ans.  153.9. 


4^2  ELEMENTS.  OF  PHYSICAL   CHEMISTRY. 

178.  The  conductivity  of  a  solution  of  AgCl  saturated 
at  18°  is  2.4X10"°;  that  of  the  water  used  is  i.i6Xio~6. 
What  is  the  solubility  of  AgCl?    ^ooKci  =  I3I-2»  ^ooAgNO8 
=  116.5,  and  J00KN03==I26-1- 

Ans.  1.02  X  io~5  moles  per  liter. 

179.  Find    the    heat    of    amalgamation    of    cadmium 
at  o°.     TT  for  a  cell  made  up  of  a  i%  Cd  amalgam  and 
mercury  in  a  solution  of  CdSC>4  is  0.06836  volts  at  o°, 
and  0.0735  at  24°-45- 

Ans.  <?=  510  cals.  per  mole  of  Cd. 
1  80.  Zn  in  a  molar  solution  of  ZnSC>4  gives  a  differ- 

ence of  potential  of  0.51  volt  and^y,=  -0.00076.   What 

is  the  heat  of  ionization  of  Zn  at  17°?       Ans.  337.4^. 

181.  Zn  +  (2lT  +  2C1')  Aq  =  (Zn"  +  2C1')  Aq  +  H2  +  342^, 
and  from  (180) 


What  is  the  heat  of  ionization  of  H  gas? 

Ans.  2H'= 

182.  A  KCN  solution  is  added  to  the  Cu  side  of  a 
Daniell  cell,  and  as  a  result  the  E.M.F.  is  zero.    What 
conclusion  is  to  be  drawn  ? 

183.  A  single  electrode   in  connection  with  the  nor- 
mal electrode  (^=0.56)  gives   an  E.M.F.   of  1.02   volts, 
the  normal  electrode  being  positive.     What  is  the  differ- 


PROBLEMS.  483 

ence  of  potential  between  the  single  electrode  and  its 
solution  ? 

Ans.  0.46  volt,  metal-  is  negative. 

184.  A    cell    with    electrodes    of    the   same    univalent 
metal  gives  an  E.M.F.  at  17°  of  0.35  volt.      The  con- 
centration  of  metal   ions   at   the   positive   electrodes   is 
0.02   mole  per  liter;    what    is    the    ionic    concentration 
on  the  other  side?       Ans.  i.637Xio~8  moles  per  liter. 

185.  With  a  cell  with  electrodes  of  a  divalent  metal, 
with  the  ionic  concentrations  on  the  two  sides  equal  to 
0.02   and   i.62Xio~8   moles  per  liter,   what  would  the 
E.M.F.  at  17°  have  been?  Ans.  0.175  volt. 

1 86.  In    a    hydrogen    gas    cell    there    is    acetic   acid 
on  one  side  and  propionic  on  the  other  of  the  same  con- 
centration;  what  is  the  E.M.F.  of  the  cell  at  17°,  and 
which  pole  is  positive? 

Ans.  Acetic  acid  positive,  ^=0.00369  volt. 

187.  In  the  above  ceU  the  pressure  of  hydrogen  gas 
is  increased  to  ten  times  its  value  on  both  electrodes. 
How  does  the  E.M.F.  change? 

1 88.  What  increase  in  E.M.F.  is  necessary  to  separate 
up  to  o.oooi  mole  of  a  univalent  metal  over  that  which 
has   separated  o.oi   mole   per   liter,  the  temperature  is 
17°?  Ans.  0.115  volt. 

189.  Assume  in  188  that  the  metal  is  divalent.     What 
would  be  the  E.M.F.  then?  Ans.  0.0575 


TABLES. 


DEGREE     OF     IONIZATION     ACCORDING    TO     CONDUC- 
TIVITY MEASUREMENTS. 
HC1 


V 

ao° 

«15° 

025° 

035° 

2 

0.899 

0.874 

0.876 

0.863 

8 

0.948 

0-937 

0.942 

0.941 

i6 

o-959 

0.946 

0.962 

0.965 

32 

0.971 

0.980 

0.980 

0.977 

128 

o-993 

0-999 

I  .OOO 

I.  OOO 

512 

1.  000 

I  .000 

1.  000 

I.  OOO 

HN03. 

V 

«0° 

«13° 

025° 

035° 

2 

0.906 

0.883 

0.879 

0.879 

8 

0.952 

0.951 

0.952 

0.950 

16 

0.982 

0-975 

0.978 

0.974 

32 

0.985 

0.988 

0.992 



128 

0.995 

o-999 

o-993 

I  .000 

512 

0.992 

o-999 

o-995 

0.999 

1024 

1.  000 

I  .000 

I  .000 

I  .000 

V 

«0° 

ffl5°4 

«25° 

«35° 

2 

o-555 

0-532 

o-547 

0.542 

8 

0.612 

0.602 

o-597 

0-572 

16 

0.643 

0.631 

0.631 

0.625 

32 

0.702 

0.685 

0.687 

0.690 

128 

0.801 

0.782 

0.817 

0.800 

512 

0.900 

0.879 

0.947 

0.888 

1024 

0.950 

0.969 

0-977 

0.984 

2048 

o-977 

0.990 

0.989 

0.996 

4096 

0.986 

I  .000 

I  .000 

0.998 

8102 

0.990 

I  .000 

-I  .  ooo 

I  .000 

485 


486  TABLES. 

DEGREE     OF     IONIZATION     ACCORDING     TO     CONDUC- 
TIVITY   MEASUREMENTS— Continued. 

KOH. 


V 

«0° 

ai%° 

«26° 

«35° 

I 

0.847 

0.803 

0.8l9 

0.833 

2 

0.884 

0.860 

0.876 

0.883 

8 

0.916 

0.907 

0.892 

0.900 

16 

0.936 

0-935 

0.938 

0  •  959 

32 

0.952 

0.958 

0.987 

0.996 

128 

0.998 

0-993 

I  .000 

i  .000 

256 

I  .000 

i  .000 

I  .000 

i  .000 

NaOH. 

i 

0.780 

0.782 

0.766 

0-774 

2 

0.838 

0.867 

0.835 

0.843 

8 

0.920 

0.904 

0.920 

o-935 

16 

0.936 

0.925 

0-944 

0.951 

32 

0.971 

0-957 

0.968 

I  .000 

128 

0.992 

0-995 

I  .000 

0.989 

256 

I  .000 

i  .000 

I  .000 

0.981 

KBr. 

V 

«0° 

«18°                      «25° 

«30° 

«35° 

2 

0.788 

0.793               0.789 

0.768 

0-747 

8 

0.823 

o  .  849         o  .  840 

0.828 

0.821 

16 

0.872 

a.88i         0.876 

0.867 

0.862 

32 

0.892 

0.921         0.900 

0.884 

0.885 

128 

0.920 

0.961        0.955 

0.938 

o  .926 

512 

0-949 

0.984        0.983 

0.987 

0.985 

1024 

I  .000 

I.  000              I.  000 

I  .OOO 

I  .000 

KI. 

V 

«o° 

ais0 

«25° 

«35° 

i 

0.767 

0-777 

0.767 

0.760 

2 

0.800 

0.817 

0.791 

0.792 

8 

0.850 

0.859 

0.852 

0.854 

16 

0.905 

0.886 

0.872 

0.881 

32 

o-935 

0.940 

0.9H 

0.930 

128 

o-979 

0.961 

0-959 

o  .  954 

512 

I  .000 

o  •  983 

0-993 

0.996 

1024 

I  .000 

0.999 

i  .000 

0.996 

TABLES.  .  487 

DEGREE     OF     IONIZATION     ACCORDING     TO     CONDUC- 
TIVITY   MEASUREMENTS.— Continued. 


2 

0.542 

o-539 

0.546 

0-532 

8 

0.648 

0.647 

0.658 

0.662 

i6 

0.709 

0.706 

0.711 

0.714 

32 

0.766 

0.740 

0.750 

0.771 

128 

0.863 

0.852 

0.893 

0.893 

512 

0.924 

0.938 

0.965 

0.949 

1024 

0.985 

0.986 

0.994 

0.989 

2048 

I  .000 

I  .000 

I  .000 

1.  000 

MOLECULAR   SURFACE   TENSION. 


HC1. 

r 

x(Mv)$ 

T             x(Mv)$ 

T 

x(Mvft 

163.1 

263-7 

175.8          244-8 

187.2 

229-3 

168.5 

255-9 

180.1          239.0 

189.9 

223.6 

171.8 

250.8 

183.2         233.6 

192.6 

221  .0 

HBr. 

181.9 

330-1 

188.9         314-6 

198.2 

294-8 

184.8 

325-6 

193-4         3°7-3 

200.5 

292.2 

186.1 

320.1 

195.3         299.6 

203.9 

283.8 

HI. 

225-3 

367.0 

230-9         355-3 

235-0 

348.0 

227.1 

362.8 

232-9         351-0 

236-5 

344-6 

229.3 

H2S. 

189.1 

349-5 

197-4         334-1 

203.9 

324.7 

191.3 

345-3 

199.7         328.3 

206.9 

316.7 

194-6 

338-0 

201.5         326.6 

210.8 

308.6 

PH3. 

167.  i 

287    2 

I7C.4           273   4 

171.  8 

*\J  (    +  A 

279-6 

*  f  3        «                      ^  /  J      T" 

179-9      265.4 

VARIATION   OF  MOLECULAR   SURFACE    TENSION    WITH 
THE  TEMPERATURE,  AND  FACTOR  OF  ASSOCIATION. 
HCl  HBr  HI  H£  PH3 


2.03          1.99          1.91          1.70 

I.O  I.O  I.I  1.4 


488    - 


TABLES. 


DATA    FOR    DIFFICULTLY 
(Bottger,  Zeit.  f.  phys. 


Salt. 

/. 

Conductivitv 
Xio6 
less  tli  at  of 
H2O. 

*a+*c 

i.e.  AOQ. 

Ionic  Concentration 
Equivalents  per  Liter. 

CaS04  

19.94 

19.68 

125.9 

•I.56XIO-2 

AgCl  

J9-95 

J-33 

125-5 

i.o6Xio-5 

AgBr  

19.96 

0.057 

127.1 

4-5  Xio-7 

AgSCN.  .  .  . 

19.96 

0.096 

116.  i 

8.2   Xio-7 

AgCN  

19.96 

0.19 

ii5-5 

1.6  Xio-8 

AgBr03.  .  .  . 

19.94 

663.9 

105-3 

6.3oXio~3 

AgI03  

iQ-95 

14.05 

92-5 

1.51X10-* 

Ag20  

19.96 

29.27 

237.2 

i.23Xio-4 

Ag20  

24.96 

35-98 

259.1 

I.38XIQ-4 

Ag2C204.  ... 

19.96 

28.76 

123-7 

2-32X  ID"4 

Ag3P04  .... 

19.46 

6.  10 

135-0 

4-5   Xio-5 

T1C1  

19.96 

1  68.0 

137-3 

1.22X10-^ 

TIBr  

20.06 

220.9 

138.9 

i.59Xio-3 

Til  

20  i  ^ 

26.18 

138.0 

i.  89X10-* 

T1SCN 

4\j  .  x^ 

10    Q4 

140  o 

j.  ^w  .  \j 

127    0 

i  .ogX  iQ-2 

TIBrO,  

V  *  V  ' 

19.94 

J.«|,W  .  V 

108.0 

^  /    y 

117.1 

9-22X  I0~3 

T1IO3 

IO    O^ 

I  ^4    2 

1  04  ^ 

I    47XIQ-3 

T12C204.... 

iy  •  yj 
19.96 

J  O'  ' 
534-4 

••vt  •  o 

135-5 

A  *  T-  /  r\  J  w 

3-94Xio-2 

T12S  

10  06 

216.0 

j.y  .  y\j 

f 

Pb"            0.0168 

PbCl2...... 

19-95 

5354 

,3 

PbCl'.        0.0109 
Cl'              0.0444 

I 

PbCl2         0.00127 

Pb"            0.0106 

PbBr2  

19.96 

3692 

134-6-j 

PbBr'        o.o  1  10 
Br'             0.0322 

I 

PbBr2        0.00114 

PbI2 

2O  .  IO 

778    4 

13^7 

2     ^  ^X  IO     ' 

oo^  •  t 

'f 

*  •  00  / 

Pb"            0.0121      ] 

Pb(SCN)2.  . 

19.96 

2640 

123.  6J 

PbSCN'    0.00727    I 
CNS'         0.00315    1 

I 

Pb(SCN)so.  000615] 

Pb(Br03)2.  . 

19.94 

4635 

112.  8 

4.  II  X  IQ— 

PbO  

19.96 

2^.  5 

244.  7 

i  .O4X  io~" 

PbC03  

V    V  w 

19.96 

3     3 

i-5 

**Tt       / 

127-5 

1.2   Xio- 

PbC204.  .  .  . 

19.96 

1.52 

I3I.2 

1.16X10- 

PbS04  
Pb3(P04)2.  .. 

24-95 

T9-95 

40.19 
0.14 

I5I.2 

142.5 

2.65X10- 
9.8  Xio- 

TABLES. 


489 


SOLUBLE   SALTS. 
Chem.,  46,  521,  1903.) 


Solubility 
Product 
(Moles). 

Total  Con- 
centration in 
Equivalents. 

a 

Grams  per  Liter. 

i  Gram  in 

X  CC. 
X 

2.44X10"* 

2.99X10- 

0.524 

2.03 

491.1 

I.I2XIO-1" 

1.06X10- 

1.  000 

1.53X10- 

653600 

2.0    XlO~lf 

4-5   Xio- 

1  .000 

0.84X10- 

11900000 

6.8  Xio-13 

8.2   Xio-° 

1.  000 

1.3   Xio- 

7300000 

2.6  Xio-12 

1.6  Xio- 

I.OOO 

2.2     XlO~ 

4540000 

3-97Xio-5 

6.7  Xio- 

0-943 

I.58XIO- 

538.8 

2.31X10-" 

i  .  54X  IQ— 

0.994 

4.35X10- 

22970 

1.52X10-" 

1.84X10- 

0.668 

2.I4XIO- 

46700 

1.93X10-" 

2.16X10- 

0.643 

2.50X10- 

40000 

6.29X10-" 

2.40X10- 

0.977 

3.65X10- 

27390 

1.3   Xio-18 

4.6  Xio- 

0.98 

O.64X  iQ— 

155000 

1.50X10-* 

1.35X10- 

0.90 

3.25X10- 

307.1 

2.53X10-' 

1.64X10- 

0.968 

0.476X10- 

2101 

3.60X10-' 

i  -92X  io~ 

0.989 

0.636X10- 

15720 

i.  i9X  IQ—  * 

1.20X10— 

0.906 

3.15X10- 

317  I 

8.50X10-* 

9.94X10- 

0.927 

3.46X10- 

288.7 

2.  19X  io~* 

1.52X10- 

0.971 

0.578X10- 

1730 

3.06X10-* 

6.35X10- 

0.62 

15.77X10- 

63.4 

o.86X  IQ— 

O  .  2  I  X  IO 

46^0 

0.044 

tv**y 

6.Q2X  IO~  2 

undiss. 

9-6iX  io~  i 

I  O4. 

PbClj 

Awir 

0.05 

4  .  54X  10  —  * 

undiss. 

8.34X10—  l 

IIQ 

PbBr2 

V 

8.ioXio~7 

2.61X10-' 

0.97 

0.470XIQ-1 

2127 

0.044 

2.78X10—* 

undiss. 

4.50X10"  1 

222.  O 

Pb(SCN)2 

3.47Xio-5 

5.78X10- 

0.72 

i3-37Xio- 

74.78 

i   <;^Xio"~ 

0.68 

i  .  7iX  IQ— 

58600 

o-36Xio-10 

i  .  3^  /\  i\j 

1.2     XlO- 

0.96 

1.6  Xio- 

595000 

o.338Xio-10 

I.22XIO- 

o-95 

1.80X10- 

550000 

1.76X10-" 

2.88X10- 

0.92 

4.38X10- 

22830 

1.2   Xio-32 

i.o   X  io~ 

0.98 

1.3  Xio- 

74OOOOO 

490 


TABLES. 


SOLUBILITY    OF  SILVER    SALTS. 


Chloracetate.. 

Borate 

Acetate. .  . 


17° 

25° 
1 8°.  6 


Propionate. .  ..   18° 
18° 


Sulphate  . 
Butyrate.  . 


25° 
18° 


6.4  Xio-2 
6  Xio~2 
5-9  Xio-2 
4.6  Xio-2 
2.33Xio-2 
2.57XIQ-2 

2.23X 


~  2 


Valerate 18° 

Benzoate 14°. 

Salicylate 15° 

Chromate 18° 

Carbonate.  ...    25° 
Sulphide 18° 


9. 5  Xio~3 
7-7Xio-3 
3.9X10-" 
i.  7  Xio-1 
1.2X10-" 
4 


SOLUBILITY   PRODUCTS. 


Sub.  / 

Hg2(SCN)2 25° 

Hg2I2 25° 

Hg2Br2 25° 

Hg2Cl2. 25° 

HgCl2 -.  ...  25° 

HgBr2 25° 

HgI2 25° 

Hg(OH)2 17° 

Zn(OH)2 17° 

Ni(OH)2 17° 

Co(OH)2 17° 

Cd(OH)2 17° 


Solubility. 

i-7Xio-7 
2.5Xio-10 
5-6XIO-8 
o.SXio--6 
0.262 


Cone,  of  Ions. 


Hg2" 
Hg" 
Hg" 
Hg" 


=  3Xio-10 
=   7Xio-8 
=   i  Xio-6 


=  2   Xio 


i.SXio-20 

I.2XIO-28 

i.3Xio-21 

3-5Xio-18 

2.6Xio-15 

8     Xio-20 

3-2Xio-29 

i.SXio-26 

4-SXio-1 

2.iXio-2 

5.4Xio-3 

2.7X10-2 


SOLUBILITIES  AT   i8c 


BaF2 

SrF2 

CaF2 

MgF2 

PbF2 : 

BaSO4 

SrSO4 

BaCrO4 

PbCr04 

BaC2O4-2H2O.  ..  . 

SrC204 

CaC2O4-2H2O.  .  .  . 
MgC2O4.2H2O.  .  .. 

ZnC2O4-2H2O 8.0 

CdC204-3H20 27.0 


/18Xioc 

•  1530 
.       172 
.         40 
.       224 

431 

2.4 
.       127 

3.2 
o.  i 

•  78-3 
54 

9.6 
200 


Mg.  Eq.  per  Liter. 
18.4 
1.87 
0.42 

2.8 

5-2 

0.020 
1.24 
0.03 
O.OOI 

o.  76 

0.52 
0.087 

5.36 

0.083 


TABLES. 


491 


HYDROLYTIC    DISSOCIATION. 


NH4N03. 

KCN. 

IT 

Per  Cent 

Per  Cent 

,, 

Per  Cent 

V 

at  100°. 

at  85°. 

at  25°. 

8 

o  .  0498 

0.0321 

i 

<    0.31 

16 

0.0776 

0.0563 

4 

0.72 

CuCl2* 

10 

I.  12 

2 

0.313 

0.252 

40 

2-34 

8 

0.407 

0.276 

Na2C03. 

32 

0.612 

0-373 

5 

2.  12 

128 

0.888 

0.5*5 

10 

3-17 

Cu(N03)2* 

20 

4-87 

8 

0.648 

0-530 

40 

7.10 

128 

1.02 

0.660 

Potassium  Phenate. 

HgCl2* 

10 

3-°5 

8 

0.6 

0.451 

50 

6.65 

•32 

1.35 

1.  06 

Borax. 

128 

2.99 

1.29 

32 

0.92 

512 

6.91 

4.87 

] 

VaC-jHgO,. 

PbCl2.* 

10 

0.008 

128 

0.768 

0.495 

128 

0.742 

0.482 

*  V  is  here  the  volume  in  liters  containing  i  equivalent  weight  in  grams. 


492 


TABLES. 


ELECTRICAL  CONDUCTIVITY   (MOLECULAR) 


Solvent. 

Salt. 

Specific  Conduc- 
tivity of  Solvent. 

Allylalcohol     

FeCl3 

6.5Xio~6 

Benzylalcohol  

FeCl3 

i.8Xio~6 

Paraldehyde 

FeClo 

<  7    4.XlO~7 

SbCl3 

Salicylaldehyde 

FeCl3 

6  oXio~6 

Furfurol  

FeCl3 

2.4Xio~5 

M^ethylpropylketone  

FeCl3 

0    qXlO~7 

Acetophenone 

FeCL 

i  8Xio~"7 

Ethylmonochloracetate  
<  < 

FeCl3 
SbCl3 
CuC)2 

<i.7Xio~6 

Ethylcyanacetate  

FeCl3 

7  ?Xio~7 

AeNOa 

<  « 

CuCL 

Ethyloxalate  

FeCl3 

7  iXio""7 

Ethylbenzoate     .            .        

FeCl3 

i  8Xio~7 

Amylnitrate                                          .  . 

FeCL 

i  8Xio~7 

Nitrobenzol  

BiClo 

<^?  t;Xio"~7 

<  < 

A1C13 

Orthonitrotoluol.  

FeCl3 

<i.8Xio~7 

SbClg 

Metanitrotoluol           

FeCl3 

<i  8Xio~7 

Benzonitril  

AgNO, 

i.pXio"6 

FeCl3 

7  t;Xio~7 

<  t 

AeNO, 

1  1 

AeCN 

« 

Pb(NO3\ 

u 

CuCl2 

1  1 

HgfCN)o 

(( 

Helo 

<  < 

AKoCfiH.Oft 

«  c 

CoCl2 

Piperidine 

AeNOo 

<i  8Xio~7 

Chinoline   

AgNO3 

i.  7X  io~7 

*  See  Lincoln.     J.  phys. 


TABLES. 
IN  NON-AQUEOUS  SOLVENTS,  25°.* 


493 


V 

P 

V 

ft 

V 

ft 

V 

n 

2O  .02 

1  7  .42 

e  -7    7  j 

2 

115.6 

2 

88.  o6 

2.62 

jo  -  /  A 
80  s    22 

6.31 

' 

°yj  -  * 

4-37 

9.81 

42.52 

18.76 

183.11 

16.51 

575-5 

16.91 

C     C7 

o  .  202 

2O    76 

O    3^6 

61.16 

o  .  522 

O     J  / 

20  .  4 

3    76 

£>\j  .  /  w 

81  .38 

Oj 
4.    71 

5-6 

45.6 

o  •  /  " 
20    78 

80.98 

T1  •    /  x 

22  .  2 

149  .  2  I 

26  .  42 

13.64 

^w  .  /«_> 
28.25 

22.83 

3I-07 

53-36 

36.98 

100.71 

42.76 

23.46 

10.28 

65-77 

n-59 

124.91 

12.03 

292.98 

13.08 

7.76 

12-45 

22.09 

13-75 

92.05 

16.38 

152-55 

17.88 

42  ^ 

O    174. 

ii  .49 

o  .  20  1 

4.4.    73 

O    337 

•  ^o 
13    32 

*-*  •  -1-  /*T 
I  .  24 

T-T-  •  /  o 

'•'  •  oo  / 

O  *  O 

15-3 

8.88 

27.08 

9.29 

44.64 

9-8 

185.22 

"•57 

10.62 

4.19 

28.3 

5-29 

58.57 

6.46 

110.97 

7.66 

29.85 

7.00 

41.67 

7-5 

59-4 

8.08 

97-77 

12.8 

13.15 

5-88 

42.29 

5-92 

94-79 

6.25 

342.85 

7-7 

20.4^ 

i   "" 

m.  28 
^ 

1.61 

CI?    21 

i  .  91 

V      TO 

21.34 

i-54 

38.74 

i-74 

j     / 

264.  16 

3.00 

644.56 

3-73 

8.5 

0.8 

34.02 

0.96 

136.07 

1.07 

272.14 

i  .  ii 

460 

3    67 

-0 

*1  .   S  I 

.  wy 

10.94 

o  •  "/ 

8-37 

25-99 

10.74 

74.28 

13-32 

201.43 

15-24 

3-4 

0.056 

6.55 

0.088 

19.82 

0.244 

39.  ic 

0-39 

10.86 

6.86 

84.77 

12-55 

448.14 

19.00 

814.8 

18.2 

2.  I 

3-37 

16.3 

S-2 

24.1 

7.66 

44.62 

10.  12 

6.1 

7.96 

24.58 

6.85 

93-7 

5-9 

159-55 

5-57 

7-55 

24.1 

5x-43 

29-5 

392-3 

40.  16 

784.6 

45-21 

14.64 

4.78 

38.4 

5-42 

100.47 

6.16 

393-o 

6.^7 

2  I  .  I 

0.8  1 

«  06 

j  .  C7 

168.5 

32^ 

o  08 

Oo  •  V" 

1.16 

o  •  ^o 

e    -j  c 

\j  .  y«_> 
O  .OI2 

I  3  .  I 

0.014 

3^7    d 

0.53 

o  •  oj 

21.8 

93-6 

1.4 

o  J  /  •  T- 
2OO  .  I 

2    7 

-  3  j 
14.64 

I  ^O  ^  .  O 

3C     3 

*  •  i 

O"T^  *    7 

OJ  -0 

74.   06 

O.2 

80?    3 

I    4.^ 

/  *T  *  wvy 

4.24 

0.368 

<J<~'o  •  o 
7.88 

0.154 

10-5 

0.091 

15.6 

0.043 

4.8 

2-45 

9.6 

2.79 

34-9 

2.8 

129.8 

3.62 

Chem.,  3.  457,  1899. 


494 


TABLES. 


SOLUBILITY    OF    GASES    IN    WATER    AT    o°,    760    MM. 
VOLUMES   OF  GAS  IN    ONE   VOLUME   OF  WATER. 


H    =0.0193 
N    =0.02035 
O    =0.04114 
NO  =  1.3052 


CO2  =       1.7967 
HgS  =       4-3706 
S02  =     79.789 
NH3=n48.8 


LIQUEFACTION   OF   GASES. 
Pressures  in  atmospheres  at  o°.       Temperatures  at  atmos.  pressure. 


S02  =   1.53 
CN  -  2.37 
HI    =  3-97 
NH3  =  4.4 
Cl      =  4 
H2S  =10 
NO  =32 
C02=35.4 
HC1=  about  42 


SO2  = 

CN    = 

NH8= 
Cl     = 


-ioc 

-22 

-36 

-50 


DIELECTRIC    CONSTANTS    AT   20°. 


Water,  H20 81.12 

Formic  acid,  HCOOH.  ...   57 
Nitromethane,  CH3NO2.  .  .    56.4 
Acetonitrile,  CH3CN.  .  . .  ab.  40 
Methylalcohol,  CH3OH. ...   32.5 
Propionitrile,  CH3CH2CN  ab.  30 
Ethylalcohol,  C2H5OH.  ...   26.8 
Acetaldehyde,  CH3COH.  . .   21.1 

Acetone,  CH3COCH3 20.7 

Glycerine, 

CH2OH.CHOH.CH2OH    16.5 
Ethylnitrate,  C2H5ONO2.  . .    19.6 

NH3 16.2 

S02 13.75 


Pyridine,  C5H5N.  ...  ab.  20 
Piperidine,  C5Hi0NH.  .  >2o 
Acetylchloride,  CHgCOCl  .  15.4 

Methyliodide,  CH31 7.2 

Ethylacetate,  CH3COOC2H5    5 . 8 

Chloroform,  CHC13 5.2 

Ether,  (C2H5)20 4.36 

Benzene,  C6H6 2.29 

Toluol,  C6H5CH3 2.31 

Aniline,  C6H5NH2 7.31 

Chinolin,  C»H7N 8.9 

Benzylcyanide,  C6H5CH2CN  15 

Benzonitrile,  C6H5CN 26 

Nitrobenzene,  C6H6NO2.  ..   36.45 


TABLES. 


495 


LOGARITHMS. 


Proportional  Parts. 


§ 

10 

11 

12 
13 
14 
15 
16 
17 
18 
19 

0 

1 

> 

A 

0128 
0531 
0899 
1239 
1553 
1847 

2122 

2380 
2625 
2856 

A 

0170 
0569 
0934 
1271 
1584 
1875 
214* 
2405 
2648 
2878 

5 

6 

7 

8 

9 

123 

456 

789 

oooo  0043 

0414  0453 
0792  0828 
H39  "73 
1461  1492 
1761  1790 
2041  2068 
2304  2330 
2553  2577 
2788  2810 

3010  3032 
3222  3243 
3424  3444 
3617  36*6 
3802  3820 
3979  3997 
4150  4166 
43144330 
4472  4487 
4624  4639 

477i:4?86 
4914  4928 
5051  5065 
5185  5198 
5315  5328 
5441  5453 
5563  5575 
5682  5694 
5798  5809 
5911  5922 

0086 
0492 
0864 
I  206 
1523 
1818 
2095 
2355 
2601 
28.^ 

0212 
0607 
0969 
1303 
l6l4 
1903 
2175 
2430 
2672 
2900 

3^18 

3324 
3522 

37  1  1 
3892 
4065 
4232 
4393 
4548 
4698 

4843 
4983 
5H9 
5250 
5378 
5502 
5623 
5740 
5855 
5966 

6075 
6180 
6284 
6385 
6484 
6580 
6675 
6767 
6857 
6946 

0253 
0645 
1004 
1335 
1644 
1931 

2201 
2455 
2695 
29^3 

0294 
0682; 
1038 
13671 
1673 
1959 
2227 
2480 
2718 
2945 

3160 
3365 
3560 

3747 
3927 
4099 
4265 
4425 
i579 
4728 

4871 
Son 
5145 
5276 
5403 
5527 
5647 
5763 
5877 
5988 

0334 
0719 
1072 
1399 
1703 
1987 
2253 
2504 
2742 
2967 

3181 
3385 
3579 
3766 
3945 
4116 
4281 
4440 
4594 
4742 

4886 
5024 
5159 
5289 
54i6 
5539 
5658 
5775 
5888 
5999 

0374 
0755 
1  106 
1430 
1732 
2014 
2279 
2529 
2765 
2989 

3201 
3404 
3598 
3784 
3962 
4133 
4298 
4456 
4609 
4757 

4812 
4811; 
3  7  10 
3  6  10 
369 
368 
35  8 
5  7 
5  7 
4  7 

17  21  2S 
15  19  23 
14  I?  21 

13  1  6  19 
12  15  18 
ii  14  17 
ii  13  16 

10  12  15 

9  12  14 
9  ii  13 

29  33  37 
26  30  34 
24  28  31 
23  26  29 

21  24  27 

2a  22  25 
18  21  24 

17  20  22 

16  19  21 
16  18  20 

20 
21 
22 
23 
24 
25 
26 
27 
28 
29 

30 
31 
32 
33 
34 
35 
36 
37 
38 
39 

40 
41 
42 
43 
44 
45 
46 
47 
48 
49 

3054 
3263 
3464 
3655 
3838 
4014 
4183 
4346 
4502 
4654 

4800 
4942 
5079 
5211 
5340 
5465 
5587 
5705 
5821 
5933 

3075 
3284 
3483 
3674 
3856 
4031 
42OO 
4362 
4518 
4669 

4814 

4955 
5092 
5224 
5353 
5478 
5599 
5717 
5832 
5944 

3096 
3.304 
3502 
3692 
3874 
4048 
4216 
4378 
4533 
4683 

3139 

3345 
354i 
3729 
3909 
4082 
4249 
4409 
4^64 
4713 

4857 
4997 

5263 
5391 
5514 
5635 
5752 
5866 
5977 

6085 
6191 
6294 
6395 
6493 
6590 
6684 
6776 
6866 
6955 

4  6 
4  J 
4  6 
4  6 

4  5. 
3  5 
3  5 
3  5 
3  5 
3  4 

8  ii  13'  15  17  ig 
8  10  I2(i4  16  18 
8  10  12,14  15  17 
7  9  11,13  IS  17 

7  9  10  12  14  15 
7  8  io!ii  13  15 
6  8  911  13  14 
6  8  9  ii  12  14 

6  7  9  10  12  13 

4829 
4969 
5105 
5237 
5366 
5490 
5611 
5729 
5843 
5955 

6064 
6170 
6274 
6375 
6474 
6571 
6665 
6758 
6848 
6937 

4900 
5038 
5172 
5302 
5428 
5551 
5670 
5786 
5899 
6010 

3  4 
3  4 
3  4 

679 
6  7  8 
678 

10  1  1  13 

10  II  I  2 
9  II  12 

3  4 
4 
4 
3 
3 
3 

568 
5  6  7 
5  6  7 
5  6  7 
5  6  7 
457 

9  10  1  1 
9  10  n 
8  10  ii 
8  9  10 
8  9  10 
8  g  10 

6021  6031 
6128  6138 
6232  6243 
6335  6345 
6435  6444 
6532  6542 
6628  6637 
6721  6730 
6812  6821 
6902^691  1 

6042 
6149 
6253 
6355 
6454 

6& 

6739 
6830 
6920 

6053 
6160 
6263 
6365 
6464 
6561 
6656 
6749 
6839 
6928 

6096 
6201 
6304 
6405 
6503 
6599 
6693 
6785 
6875 
6964 

6107 
6212 
6314 
6415 
6513 
6609 
6702 
6794 
6884 
6972 

6117 
6222 
6325 
6425 
6522 
6618 
6712 
6803 
6893 
6981 

7067 
7152 
7235 
73i6 
739<5 
7474 
7551 
7627 
7701 
7774 

3 

I 

3 

^ 

4  5  6 
4  5  6 
4  5  6 
4  5  6 
4  5  6 
456 

8  9  10 
7  8  9 
7  8  9 
7  8  9 
789 
7  8  9 

3 

! 

455 
445 
445 

678 
6  7  8 
678 

50 
51 
52 
53 
54 
55 
56 
57 
58 
59 

6990  6998 
7076:7084 
7160  7168 
7243*7251 
7324  7332 
7404  7412 
7482  7490 
7559  7566 
7634  7642 
7709  7716 

7007 
7093 
7177 
7259 
7340 
7419 

7497 
7574 
7649 
7723 

7016 
7101 
7i85 
7267 
7348 
7427 
7505 
7582 
7057 
7731 

7024 
7110 
7193 
7275 
7356 
7435 
7513 
7589 
7664 
7738 

7033 
7118 
7202 
7284 
7364 
7443 
7520 
7597 
7672 
7745 

7042 
7126 
7210 
7292 
7372 
7451 
7528 
7604 
7679 
7752 

7050 
7135 
7218 
7300 
738o 
7459 
75  -< 
7612 
7686 
7760 

7059 
7143 
7226 
73°8 
7388 
7466 
7543 
7619 
7694 
7767 

3 

2 

2 

1  i 

2 

: 

345 
3  4 
3  4 
3  4 
3  4 
3  4 
3  4 
3  4 
3  4 
3  4 

678 
678 
6  7  7 
667 
667 
5  6  7 

5  \  7 
5  6  7 

5  6  7 
5  6  7 

496 


TABLES. 
LOGARITHMS— Continued. 


Proportional  Parts. 


j 

60 
61 
62 
63 
64 
65 
66 
67 
68 
69 

0 

7782 
7853 
7924 
7993 
8062 
8129 
8i95 
8261 
8325 
8388 

1 

2 

3 

4 

.|. 

7       8   !    9 

1         1 

123 

456 

789 

566 
566 
566 
556 
5  5  6 
5  5  6 
556 
5  5  6 
456 
4  5  6 

7789 
7860 
7931 
8000 
8069 
8136 
8202 
8267 
8331 
8395 

7796 
7868 
7938 
8007 
8075 
8142 
8209 
8274 
8338 
^40  1 

7803 
7875 
7945 
8014 
8082 
8149 
8215 
8280 
8344 
8407 

7810 
7882 
7952 
8021 
8089 
8156 
8222 
8287 
8351 
8414 

7818  7825  7832^839  7846 
7889  7896  79°3  7910  7917 
7959  79667973  7980  7987 
8028  8035  8041  8048  8055 
8096  8102  8109  8116  8122 
8162  81698176  8182  8189 
8228  8235  8241^8248  8254 
8293  8299(83o6  8312  8319 
8357,8363,8370  83768382 
8420^426843284398445 

344 
344 
334 
334 
334 
334 
334 
334 
334 
234 

70 
71 
72 
73 

74 
75 
76 

77 
78 
79 

80 
81 

82 
83 
84 
85 
86 
87 
88 
89 

90 
91 
92 
93 
94 
95 
96 
97 
98 
99 

8451 
8513 
8573 
8633 
8692 
875i 
8808 
8865 
8921 
8976 

8457 
8519 
8579 
8639 
8698 
8756 
8814 
8871 
8927 
8982 

9036 
9090 
9143 
9196 
9248 
9299 
9350 
9400 
9450 
9499 

8463 
8525 
8585 
8645 
8704 
8762 
8820 
8876 
8932 
8987 

8470 
853T 
8591 
86<;i 
8710 
8768 
8825 
8882 
8938 
8.993 

8476 
8S17 
8597 
8657 
8716 
8774 
8831 
8887 
8943 
8998 

8482^488  8494'8soo  8506 
8S438S49!8SSS  8561  8567 
8603  8609  8615  8621  8627 
8663  8669  8675  8681  8686 
8722  8727  8733  8739  8745 
87798785(8791  8797  8802 
8837  8842:884888548859 
8893^899  8904  8910  8915 
8949^954  8960  8965  8971 
9004  9009  9015  9020  9025 

3     4 
3     4 
3     4 
3     4 
3     4 

4  5  6 

455 
455 
455 
455 

3     3 
3     3 
3     3 
3     3 

455 
445 
445 
445 

445 
445 

445 

9031 
9085 
9138 
9191 
9243 
9294 
9345 
9395 
9445 
9494 

9042 
9096 
9149 
9201 
9253 
9304 
9355 
9405 
9455 
9504 

9047 
9101 
9i54 
9206 
9258 
9309 
936o 
9410 
9460 
9509 

9557 
9605 
9652 
9699 
9745 
9791 
9836 
9881 
9926 
9969 

9053 
9106 
9159 
9212 
9263 
9315 
9365 
9415 
9465 
9513 

9562 
9609 
9657 
9703 
975° 
9795 
9841 
9886 
9930 
9974 

905819063 
9H2|9ii7 
9165  9170 
9217  9222 
9269  9274 
9320^325 
9370,9375 
9420,9425 
9469  9474 
95189523 

9069  9074  9079 
9122  9128  9133 
9175  9180  9186 
9227  9232  9238 
9279  9284  9289 
9330:9335  9340 

I  I 

3     3 

3     3 
3     3 

445 
4  4  5 
445 
344 
344 
344 

9430  9435  9440 
9479  9484  9489 
9528I9533  9538 

0 

o 

0 

3 
3 
3 

9542 
959° 
9638 
9685 
9731 
9777 
9823 
9868 
9912 
9956 

9547 
9595 
9643 
9689 
9736 
9782 
9827 
9872 
9917 
996i 

9552 
9600 
9647 
9694 
974i 
9786 
9832 
9877 
992i 
9965 

9566  9571 
9614  9619 
9661  9666 
9708^713 
9754'9759 
980019805 
9845  9850 
9890  9894 
9934  9939 
99789983 

9576  9581  9586 
9624  9628  9633 
9671  9675  9680 
9717  9722  9727 
9763  9768  977: 
9809  9814  9818 
9854  9859  9863 
9899,9903  9908 
9943  9948  9952 
99879991  9996 

0 
0 

o 

0 
0 

o 

0 

o 

0 
0 

3 
3 
3 
3 
3 
3 
3 
3 
3 
3 

344 
344 
344 
3  4  4 
344 
344 
344 
344 
344 

APPENDIX. 


THE   OSMOTIC    PRESSURES    AND    FREEZING-POINTS 
OF   SOLUTIONS   OF   CANE-SUGAR. 

In  a  paper  which  has  just  appeared  (July),  Morse  and 
Frazer  (Am.  Chem.  Jour.,  34,  i,  1905)  give  their  results 
for  the  osmotic  pressures  and  freezing-points  of  cane- 
sugar  solutions  of  varying  concentration.  These  results 
show  that  the  van't  Hoff  law  (p.  134)  holds  throughout 
for  solutions  of  sugar,  ij  the  volume  of  ike  pure  solvent 
is  substituted,  in  the  law,  jor  that  oj  the  solution.  In 
their  words:  "When  we  dissolved  a  gram-molecular 
weight  of  cane-sugar  (342.22  grams')  in  1000  grams  of 
water,  i.e.,  in  that  mass  of  solvent  which  Jtas  the  unit 
volume,  i  liter,  at  the  temperature  of  maximum  density — 
we  found  its  osmotic  pressure,  at  about  20°,  in  quite  close 
accord  with  the  pressure  which  a  gram-molecular  weight 
oj  hydrogen  would  exert,  at  the  same  temperature,  ij  its 
volume  were  reduced  to  i  liter,  i.e.,  to  that  volume  which 
the  unit  mass  oj  solvent  has  at  the  temperature  oj  greatest 
density. " 

497 


49^  APPENDIX. 

For  cane-sugar,  then,  and  presumably  for  all  other 
substances,  we  can  express  the  law  as  follows:  A  sub- 
stance in  solution  "exerts  an  osmotic  pressure  equal  to 
thai  which  it  would  exert  if  it  were  gasified  at  the  same 
temperature  and  the  volume  oj  the  gas  were  reduced  to 
that  oj  the  solvent  in  the  pure  state."  Or,  in  other 
words,  it  "exerts  an  osmotic  pressure  throughout  the 
larger  volume  oj  the  solution  equal  to  that  which  as  a  gas 
it  would  exert  ij  confined  to  the  smaller  volume  oj  the  pure 
solvent" 

It  will  be  observed  here  that  the  two  forms  of  the  law 
will  only  differ  for  concentiated  solutions,  and  the  stronger 
the  solution,  the  greater  will  be  the  difference.  So  long 
as  the  amount  of  substance  dissolved  is  relatively  small, 
the  volume  of  the  solution  will  be  practically  equal  to 
that  of  the  pure  solvent,  and  no  difference  in  osmotic 
pressure  will  be  observed.  As  the  amount  increases, 
however,  the  difference  between  the  volume  of  solvent 
and  solution  will  become  greater  and  greater,  and  the 
osmotic  pressures,  according  to  the  two  laws,  will  diverge 
more  and  more.  Our  definition  of  molecular  weight 
in  solution  (p.  135)  should  then  be  slightly  altered  in 
that  the  22.4  liters  should  be  the  volume  of  the  pure 
solvent  and  not  that  of  the  solution.  The  error  in  the 
first  portion  of  the  definition,  however,  will  be  exceed- 
ingly slight,  but  will  increase  as  the  volume  decreases, 
so  that  the  second  part,  when  it  refers  to  a  smaller 


APPENDIX. 


499 


volume   than  22.4   liters,   can  only  be  used   when  the 
volume  taken  is  that  of  the  pure  solvent. 

In  the  light  of  this  new  form  of  the  van't  Hoff  law 
the  abnormal  results,  mentioned  on  page  135,  become 
perfectly  normal,  as  a  glance  at  the  following  table  will 
show. 

OSMOTIC  PRESSURES  AND  MOLECULAR  WEIGHTS. 
SUGAR  SOLUTIONS  AT  ABOUT  20°. 


Weight- 
normal. 
Moles  in  1000 
gr.  H2O 

(wf. 

Volume- 
normal. 
Moles  per 
Liter. 

Pressures  at  Same 
Temperature.         ^-  = 

Gaseous.    Osmotic  (P). 

P 

0.05 

o  .  04948 

I.  21 

1.26 

327-5 

0.10 

0.09794 

2.40 

2.44 

336.9 

O.2O 

o.  19192 

4.82 

4.78 

345-2 

0.25 

0.23748 

6.06 

6.05 

342-9 

0.30 

0.28213 

7.22 

7-23 

342.0 

O.4O 

0.36886 

9.68 

9.66 

343-1 

0.50 

0.45228 

12.07 

12.09 

341-7 

O.6o 

0-53252 

14.58 

14.38 

347-1 

0.70 

0.60981 

17.16 

17.03 

344-8 

0.80 

o  .  68428 

19.17 

19.38 

338.5 

0.89101 

o  .  75000 

21.48 

21.21 

346.5 

0.90 

0.75610 

21.73 

21.  8l     ' 

340.0 

I  .00 

0.82534 

24.27 

24-49 

339-2 

Mean.  .  .   341.2 

These  are  but  a  few  of  the  many  values  determined  by 
the  authors  at  these  concentrations.  The  slight  varia- 
tions in  the  calculated  and  observed  values  they  assume 
to  be  due  to  experimental  errors  which  will  be  eliminated 
in  future  measurements.  The  law  in  this  form,  then, 
is  to  be  regarded  as  holding  accurately  for  solutions 
between  these  limits.  Whether  stronger  solutions  will 


Soo  APPENDIX. 

show  abnormal  pressures,  as  do  gases,  is  not  as  yet 
known,  but  further  work  for  this  purpose  will  undoubtedly 
be  undertaken  in  the  near  future. 

The  authors  have  also  studied  the  freezing-points  of 
sugar  solutions.  The  laws  already  mentioned  above 
(pp.  176-187)  have  been  found  for  some  unknown  reason 
to  be  unsuited  to  strong  sugar  solutions,  but  by  the 
following  relation,  found  by  them,  the  freezing-point 
may  be  calculated  for  all  concentrations.  Calling  IF  the 
concentration  of  the  solution  according  to  the  weight- 
normal  standard,  i.e.,  the  fraction  of  a  mole  of  sugar 
which  is  dissolved  in  1000  grams  of  water,  D  the  density 
of  the  solution  at  its  freezing-point,  i.e.,  the  relative 
weights  of  equal  volumes  of  solution  and  pure  solvent 
near  the  temperature  of  freezing,  and  J  the  depression 
of  the  freezing-point,  they  find 

J  =  i.S$WD  for  all  concentrations, 
where  1.85  is  the  molecular  depression  of  the  freezing- 

0.02  T2 
point  of  water,  as  calculated  from  the  formula  K  = 

(p.  179).  For  dilute  solutions  this  relation  is  just  what 
we  found  it  to  be  above,  for  the  density  D  in  such  a 
case  will  be  practically  unity.  For  strong  solutions, 
however,  J,  as  calculated,  will  be  larger  than  that  by 
the  simpler  relation.  The  agreement  between  the  cal- 
culated and  observed  values  of  J  is  exceedingly  close 
as  may  be  seen  from  the  results  below. 


APPENDIX.  501 

FREEZING -POINTS  OF  SUGAR  SOLUTIONS. 
W.  D.  A  Calc.  A  Obs. 


O.  10 
O.20 
0.30 
0.40 
0.50 
0.6o 
O.7O 
0.80 
O.OXD 
I  .00 


.0129  o°.i87  o°.i87 

.0257  o°.379  o°.373 

.0380  o°.576  o°.574 

.0497  o°-777  o°-776 

.0611  o°.98i  o°.97o 

.0717  i°.i89  i°.i87 

.0825  i°.4oi  i°-398 

.1016  i°-834  i°-837 


.1110       2°. 060       2°. 082 
1.26806       1-1340       2°. 660       2°. 660 

"The  last  solution  (the  1.26806  weight-normal  one)  is 
a  volume-normal  solution,  i.e.,  a  solution  containing  a 
gram-molecular  weight  of  sugar  in  a  liter  volume.  The 
depression  of  the  freezing-point  of  this  solution  has 
been  regarded  as  abnormal  by  the  difference  between 
i°.85  and  2°.66,  but  it  will  be  noticed  that  it  conforms 
to  the  rule  with  the  same  degree  of  precision  as  the  o.i 
weight-normal  solution."  The  deviations  for  the  0.5 
and  i.o  weight-normal  solutions  may  be  due  to  experi- 
mental errors,  according  to  the  authors,  for  the  concen- 
trations on  either  side  show  close  conformity  to  the  rule. 

The  authors  also  find  a  relation  existing  between  the 
freezing-point  and  the  osmotic  pressure  of  a  solution  of 
sugar.  Calling  P  the  theoretical  osmotic  pressure  at 
o°,  D,  as  before,  the  density,  and  J  the  observed  depres- 
sion of  the  freezing-point,  then  -^  =0.082.  From  this, 
of  course,  knowing  either  A  or  P  we  can  calculate  P  or 


502 


APPENDIX. 


J.     The  application  of  this  relation  for  the  two  purposes 
is  shown  in  the  tables  below. 

FREEZING-POINTS  OF  SUGAR  SOLUTIONS. 


Concentra- 
tion     Weight- 
normal. 

Osmotic 
Pressure 
at  o°.     At- 
mospheres. 

Density  of 
Solution  at 
its  Freezing- 
point. 

Observed 
Lowering  of 
Freezing- 
point. 

Osmotic 
Pressure 
at  o°  X  Den- 
sity X  0.082. 

O.I 

2.25 

1.0129 

o°.i87 

0°.l87 

0.2 

4-5° 

1.0257 

o°.373 

o°.378 

0-3 

6.75 

1.0380 

o°-574 

o°-574 

0.4 

9.00 

1.0497 

o°.776 

o°.774 

o-5 

11.24 

I  .0611 

o°.969 

o°.978 

0.6 

13-49 

1.0717 

i°.i87 

i°.i8s 

0.7 

15-74 

1.0825 

i°.398 

i°-397 

0.8 

17.99 

1.0918 

I°.6l2 

i°.6io 

0.9 

20.24 

i  .  1016 

i°.837 

i°.828 

I.O 

22.49* 

I  .  I  I  10 

2°.  082 

2°.  049 

I  .  26806 

28.76 

1.1340 

2°.  660 

2°.  660 

OSMOTIC  PRESSURES  OF  SUGAR  SOLUTIONS. 


Values  of 


Concentra- 
tion.     Weight- 
normal. 

Observed 
Lowering  of 
Freezing- 
point. 

Density  of 
Solution  at 
its  Freezing- 
point. 

Theoretical 
Gas  or 
Osmotic 
Pressure 
at  o°. 

0.082/7'  '  '' 
Calculated 
Osmotic 
Pressure. 
Atmospheres. 

0.  I 

o°.i87 

I  .0129 

2.25 

2.25 

0.2 

o°-373 

1.0257 

4-50 

4-43 

0-3 

o°-574 

I  .  0380 

6-75 

6-75 

0.4 

o0.776 

1.0497 

9.00 

9.01 

0-5 

o°.969 

I  .0611 

11.24 

11.13 

0.6 

i°.i87 

1.0717 

13-49 

13-5° 

0.7 

i°.398 

1.0825 

15-74 

J5-75 

0.8 

I°.6l2 

I  .0918 

17.99 

18.00 

0.9 

i°.837 

i  .  1016 

20.24 

20.33 

I  .0 

2°.  082 

I  .  IIIO 

22.49* 

22.85 

i  .  26806 

2°.  660 

1.1340 

28.76 

28.82 

The  pressures  marked  with  an  asterisk 
considered  as  22.4  atmospheres,  for  which 


(*)  are  those  which  we  have 
the  authors  use  22.488. 


INDEX. 


Adiabatic  compression  and  expansion 45-48 

Ampere 364 

Analytical  reactions 352-363 

Association,  Factor  of 81-85 

Atomic  and  combining  weights 6-8 

Atoms  in  molecule,  Number  of 49 

Avogadro's  law 15 

Babo,  Law  of 153 

Basicity  of  an  acid 386-387 

Boiling-point 65-68 

,  Determination  of  the '174 

,  Increase  of  the 169 

,  Molecular  increase  of  the 169 

Boyle's  law 9 

Calorie,  Large 202 

,  Ostwald 202 

,  Small 202 

Capacity  factor  of  energy 3 

Catalytic  action  of  hydrogen  ions 268-272 

Cell,  Bichromate 437~438 

,  Chemical  or  thermodynamical  theory  of  the 402-405 

,  Clark 435-436 

,  Concentration 422-427 

,  Daniell 401,  402,  404 

,  Gas 404,  424,  431 


504  INDEX. 

PAGE 

Cell,  Helmholtz 400 

,  Leclanche" 436 

,  One-volt 400 

,  Osmotic  theory  of  the 404,  414 

,  Storage 438,  439 

,  Weston 400 

Charles,  Law  of 9 

Chemical  change 222 

at  constant  pressure 208 

volume 207 

kinetics 261 

Coefficient  of  affinity 283 

partition  or  distribution 188-189,  251 

Color  of  solutions 346 

Conductivity,  Determination  of  the 376 

of  mixtures 396 

neutral  salts 387 

organic  acids 381 

,  Solubility  by  aid  of  the 39I-392 

,  Temperature  coefficient  of 390 

,  The  molecular  and  equivalent 375 

,  The  specific 374 

Constant,  Dissociation 283,  296 

lonization 296 

of  a  reaction  of  the  ist  order,  The 265 

2d      "         "     272 

3d      "         "     276 

hydrolytic  dissociation,  The 32 1 

Coulomb 364 

Cycle,  The 54 

Dalton's  law 13 

Decomposition  of  H2O,  Primary 447 

values 440-442 

Dielectric  constant  and  dissociating  power 393 

Dissociation  and  external  pressure    33 

,  Constant  of 233-264,  281-306 

,  Degree  of .    23,  146,  283,  297 

,  Electrolytic ,> 146 

from  E.M.F 427 


INDEX.  505 

PAGE 

Dissociation,  Gaseous 23,  233 

,  Heat  of,  of  solids 255 

gases. 259 

,  Hydrolytic 316-329 

solution,  Non-electrolytic 250 

of  double  molecules  in  solution 251 

H20 313,  388,  430-432 

polyatomic  molecules 26-28 

Dissociating  power  and  dielectric  constant 393 

Dilution  law,  Bancroft's 297 

Ostwald's 282 

Rudolphi's 296 

van't  Hoff's 296 

Distribution  of  a  substance  between  two  solvents 188-189,  251 

Dyne 2 

Electric  conductivity,  see  Conductivity. 

Electricity,  Mechanical  equivalent  of 365 

,  Thermal  equivalent  of 365 

Electrochemistry 364-452 

Electrodes,  Concentration  of 422-426 

,  Normal 417-418 

Electrolysis 440 

Electrolytic  solution -pressure 409,  433 

Electromotive  force 399-440 

Energy,  Electrical 5,  365 

,  Factors  of 3 

,  Factors  of  heat 59 

,  Kinetic 5 

,  Mechanical 2 

,  Potential 2 

,  Volume 4 

Entropy 59 

Equation  of  equilibrium 227,  282,  296,  297 

state  for  gases 12 

van  der  Waals 35 

Equilibrium 222 

and  external  pressure 33 

between  gases  and  solids 241-245 

,  Constant  of 228 


506  INDEX. 


Equilibrium,  Effect  of  temperature  upon 252 

Erg 2 

Evaporation,  Heat  of • 68-76 

Faraday,  Law  of 366 

Film,  Semipermeable. 128 

Force 2 

Freezing-point  and  external  pressure 93  -96 

vapor-pressure 180-181 

,  Depression  of  the ...    176 

,  Determination  of  the 184-185 

,  Molecular  depression  of  the 176 

of  mixtures 396 

Fusion,  Latent  heat  of .    .   93-103 

Gas  constant,  Molecular 12 

,  Specific ii 

,  Equation  of  state  for  a 12 

laws 9-1 5 

Gases  and  liquids,  Distinction  between 61 

in  liquids.  .  . 116-1.18 

,  Specific  gravity  of 15 

.  Determination  of  specific  gravity  of 16-21 

Gay-Lussac,  Law  of 9 

Heat,  Atomic 9J~93 

,  Capacity  for 39 

,  Determination  of  specific S°-52 

,  Latent 68-69,  72~?6,  98-103 

of  association  of  the  ions  of  water. 216 

dissociation 255-256,  259-261,  421 

evaporation 69,  72,  254 

formation 2°5>  206 

ionization.  c 301,  421 

neutralization 216,  304 

precipitation 218-219 

solution 203 

vaporization .    68-76,  256-260 

,  Specific 38,  S° 

Hess,  Law  of 200 


INDEX.  507 

PAGE 

van't  Hoff,  Law  of 134 

Hydrolytic  dissociation,  Constant  of 321 

Indicators,  Action  of 348 

Intensity  factor 2 

Ionic  concentrations^ 426 

equilibria 281,  334-348 

lonization 139-152,  281-316 

and  solubility 305-316 

constant 283-301 

from  increased  solubility 329~334 

Ions 139-152 

and  analytical  chemistry 352-363 

in  thermal  reactions 214-221 

,  Migration  of  the 364-374 

Isohydric  solutions 291 

Joule 203 

Kohlrausch,  Law  of 378 

Liquids  in  liquids . '. 1 18-124 

Liter  atmosphere , 13,  41 

Mariotte's  Law 9 

Mass  action,  Law  oi.  .  .  .^ 228 

Migration  of  the  ions,  Absolute  velocity  of 385,  386 

in  terms  of  conductivity 381 

Mixtures,  Conductivity  of 396 

,  Freezing-point  of 396 

Mole 12 

Molecular  weight  from  depressed  solubility. 189-191 

depression  of  freezing-point 176-185 

depression  of  vapor-pressure 154 

distribution  between  two  solvents 251 

heat  of  evaporation 69 

increase  of  the  boiling-point 118 

osmotic  pressure 135 

surface  tension 79~&5 

vapor  density 16 

law  of  Young  &  Thomas 70 


508  INDEX. 

PAGE 

Non-homogeneous  systems 241,  279 

Ohm 364 

Osmotic  pressure  and  molecular  weight 135 

vapor-pressure 161-168 

,  Simple  demonstration  of 137 

theory  of  the  cell 404,  414 

Partition,  Coefficient  of 188-189,  251 

Perpetual  motion  of  the  ist  kind 41 

2d      "     53 

Phase  rule . . . ; 104-115 

Physical  chemistry i 

Point,  Transition in 

,  Triple in 

Polarization,  Theory  of 442-447 

Pressure,  Critical.  . ,. 63 

,  Electrolytic  solution 409,  433 

,  Molecular  depression  of  the  vapor 154 

,  Osmotic 128^138 

,  Relative  depression  of  the  vapor 154 

,  Solution 126,  152 

,  Vapor 153,  161,  180 

Problems 453-483 

Raoult,  Law  of 154 

Reactions  between  solids  and  liquids 279 

,  Incomplete , 278 

of  the  ist  order.  ...     ...  265 

2d       "    .  = 272 

3d       "   276 

,  Reversible 222 

Separation  of  metals  by  graded  E.M.F.'s 449 

Solid  state,  The 91-103 

solutions 192 

Solids  in  liquids 124 

Solution,  Heat  of ...   203, 257 


INDEX.  5°9 

PAGE 

Solution  pressure  of  metals,  Electrolytic 409,  433 

Solutions 116 

,  Color  of 346 

,  Isohydric 291 

,  Saturated 125-128 

,  Solid 192 

,  Vapor-pressure  of 153, 161 

Solubility,  Depressed 189-190,  308-313 

,  Increased 311 

from  conductivity 39i~393 

E.M.F 429-430 

,  Law  of 312 

of  sulfids,  Law  of 358 

Specific  gravity  in  gaseous  form 13-21 

Speed  of  reaction  and  temperature 280 

Sublimation,  Heat  of 78-80, 102 

Sugar,  Inversion  of 265-268 

Surface-tension 78-85 

and  critical  temperature 86-96 

molecular  weight 78-85 

Tables 485-496 

Temperature,  Absolute 10 

,  Critical 63-65,  70 

Thermochemistry 200 

,  Definition  of 20 

Thermodynamics,  ist  principle  of 38 

2d          "        " 52-59 

Trouton,  Law  of 69 

Units,  Electrical 364 

Van  der  Waals,  equation  of 35 

Vapor-densities,  Abnormal 21-29 

Vapor-pressur?,  see  Pressure. 

Volume,  Critical 63-65,  70 

,  Molecular •„ 12,  80 

Volt 364 


5io  INDEX. 

PAGE 

Water,  lonization  of r  -  -  -          313,  388,  43Q-432 

Work ^ 

,  Maximum 2 

Wiillner,  Law  of 153 

Zero,  Absolute 10 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 
BERKELEY 

Return  to  desk  from  which  borrowed. 
This  book  is  DUE  on  the  last  date  stamped  below. 


APft  8  1! 

26Mar'57Kl 
REC'D  LI 

APR  12 


LD  21-100m-9,'47(A5702sl6)476 


YB   16909 


<} 


UNIVERSITY  OF  CALIFORNIA  LIBRARY 


